Category: OP Malhotra Solutions

Interactive S Chand ISC Maths Class 12 Solutions Chapter 8 Differentiation Ex 8(a) engage students in active learning and exploration.

S Chand Class 12 ICSE Maths Solutions Chapter 8 Differentiation Ex 8(a)

Differentiate with respect to x :

Question 1.
(i) x⁵
(ii) 6x⁸
(iii) x^3/4
(iv) 4\(\sqrt{x}\)
(v) 8x^-3/4
(vi) \(\sqrt{x^3}\)
(vii) \(\frac{9}{x}\)
(viii) \(\frac{7}{x^2}\)
Solution:
(i) Let y = x⁵,
Diff both sides w.r.t x, we have
\(\frac{d y}{d x}=\frac{d}{d x} x^5=5 x^4\)

(ii) Let y = 6x⁸,
Diff both sides w.r.t x, we have
\(\frac{d y}{d x}=\frac{d}{d x}\left(6 x^8\right)=48 x^7\)

(iii) Let y = x^3/4
Diff both sides w.r.t x, we have
\(\frac{d y}{d x}=\frac{d}{d x} x^{3 / 4}=\frac{3}{4} x^{\frac{3}{4}-1}=\frac{3}{4} x^{\frac{-1}{4}}\)

(iv) Let y = 4\(\sqrt{x}\)
Diff both sides w.r.t x, we have
\(\frac{d y}{d x}=\frac{d}{d x} 4 \sqrt{x}=4 \frac{d}{d x} x^{1 / 2}\)
= \(4 \times \frac{1}{2} x^{\frac{1}{2}-1}=\frac{2}{\sqrt{x}}\)

(v) Let y = 8x^-3/4 ;
Diff both sides w.r.t x, we have
\(\frac{d y}{d x}=8\left(\frac{-3}{4}\right) x^{\frac{-3}{4}-1}=-6 x^{-7 / 4}\)

(vi) Let y = \(\sqrt{x^3}=x^{3 / 2}\)
Diff both sides w.r.t x, we have
\(\frac{d y}{d x}=\frac{d}{d x} x^{3 / 2}=\frac{3}{2} x^{\frac{3}{2}-1}=\frac{3}{2} \sqrt{x}\)

(vii) Let y = \(\frac{9}{x}\)
Diff both sides w.r.t x, we have
\(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{9}{x}\right)=9 \frac{d}{d x} \frac{1}{x}\)
= 9\(\frac{d}{d x} x^{-1}=9(-1) x^{-1-1}\)
= \(\frac{-9}{x^2}\)

(viii) Let y = \(\frac{7}{x^2}\) ;
Diff both sides w.r.t x, we have
\(\frac{d y}{d x}=7 \frac{d}{d x} x^{-2}=7(-2) x^{-2-1}=\frac{-14}{x^3}\)

Question 2.
(i) (2x + 3)⁵
(ii) (1 – x)⁴
(iii) \(\sqrt{8-7 x}\)
(iv) (3x² + 5)⁹
Solution:

Question 3.
(i) \(\frac{2}{x}+\frac{1}{\sqrt{x}}\)
(ii) (2x – 1) (3x + 2)
(iii) \(x^4-2 x+\frac{1}{x^2}\)
(iv) 2x² (x + 1) + 2
(v) \(\frac{3 x^4-x}{x^3}\)
Solution:

Question 4.
(i) cos 7x
(ii) tan ax
(iii) sec 9x
(iv) sin x²
(v) cos \(\sqrt{x}\)
(vi) 2 cosec bx²
Solution:
(i) Let y = cos 7x ;
Diff both sides w.r.t x, we have
\(\frac{d y}{d x}=\frac{d}{d x} \cos 7 x\)
= sin 7x \(\frac { d }{ dx }\) (7x) = – 7sin 7x

(ii) Let y = tan ax ;
Diff both sides w.r.t x, we have
\(\frac{d y}{d x}=\sec ^2 a x \frac{d}{d x} a x\) = a sec²ax

(iii) Let y = sec 9x ;
Diff both sides w.r.t x, we have
\(\frac{d y}{d x}=\frac{d}{d x}\)sec 9x = sec9x tan9x \(\frac { d }{ dx }\) 9x
= 9 sec 9x tan 9x

(iv) Let y = sin x² ;
Diff both sides w.r.t x, we have
\(\frac{d y}{d x}=\cos x^2 \frac{d}{d x} x^2=2 x \cos x^2\)

(v) Let y = cos \(\sqrt{x}\) ;
Diff both sides w.r.t x, we have
\(\frac{d y}{d x}=-\sin \sqrt{x} \frac{d}{d x} x^{\frac{1}{2}}\)
= – sin \(\sqrt{x} \frac{1}{2} x^{\frac{-1}{2}}\)
= – \(\frac{\sin \sqrt{x}}{2 \sqrt{x}}\)

(vi) Let y = 2 cosec bx³ ;
Diff both sides w.r.t A, we have
\(\frac { dy }{ dx }\) = – 2 cosec bx³ cot bx³ \(\frac { d }{ dx }\) bx³
= – 6bx² cosec bx³ cot bx³

Question 5.
(i) \(\frac{x^3}{3 x-2}\)
(ii) \(\frac{x}{\sin x}\)
(iii) \(\frac{1+\cos x}{1-\cos x}\)
(iv) \(x^2-\sqrt{(1+x)}\)
(v) sin 2x cos² x
(vi) tan⁴ 7x
(vii) \(\frac{1}{\sin x+\cos x}\)
(viii) \(\frac{1+\cos x}{x}\)
Solution:

Question 6.
Given that y = \(\frac{\sin x-\cos x}{\sin x+\cos x}\), show that \(\frac{d y}{d x}=1+y^2\)
Solution:

Question 7.
Differentiate with respect to x :
(i) (2x² – 1) (x³ + 4)³
(ii) \(\frac{x^4+1}{\sqrt{1+x}}\)
(iii) \(\left(x+\frac{1}{x}\right)^{-1}\)
(iv) tan⁴2x
Solution:

Question 8.
Find the gradient function \(\frac { dy }{ dx }\) for each of the following :
(i) y = x – 7x²
(ii) y = 4x⁷ – 3x³ + 5x – 11.
Solution:
(i) Given y = x – 7x² ;
Diff both sides w.r.t x, we have
\(\frac { dy }{ dx }\) = 1 – 14x

(ii) Given y = 4x⁷ – 3x³ + 5x – 11
Diff both sides w.r.t x, we have
\(\frac { dy }{ dx }\) = 28x⁶ – 9x² + 5

Question 9.
Find the gradients of the following curves at the points indicated.
(i) y = x² + 5x at (0, 0)
(ii) y = (x + 1) (2x + 3) at (2, 21)
(iii) y = 2x² – x + \(\frac { 4 }{ x }\) at (2, 8)
Solution:
(i) Given y = x² + 5x ;
Diff both sides w.r.t x, we have
\(\frac { dy }{ dx }\) = 2x + 5
at (0, 0); \(\frac { dy }{ dx }\) = 2 x o + 5 = 5

(ii) Given y = (x + 1) (2x + 3) = 2x² + 5x + 3
Diff both sides w.r.t x, we have
\(\frac { dy }{ dx }\) = 4x + 5
∴ at (2, 21); \(\frac { dy }{ dx }\) = 4 x 2 + 5 = 13

(iii) Given y = 2x² – x + \(\frac { 4 }{ x }\)
Diff both sides w.r.t x, we have
at (2, 8); \(\frac { dy }{ dx }\) = 8 – 1 – \(\frac { 4 }{ 4 }\) = 8 – 1 – 1 = 6

Question 10.
If f (x) = 3x² – 4x, find the value of a given that f ‘(a) = 5.
Solution:
Given f (x) = 3x² – 4x
Diff both sides w.r.t x, we have f'(x) = 6x – 4
∴ f ‘(a) = 6a – 4
Also f ‘(a) = 5
∴ 5 = 6a – 4 ⇒ 6a = 9 ⇒ a = \(\frac { 3 }{ 2 }\)

Question 11.
Differentiate from first principle.
(i) 3x
(ii) (x + 1) (2x – 3)
(iii) \(\frac{2-x}{4+3 x}\)
(iv) x^-3/4
(v) \(\sqrt{x}+\frac{1}{\sqrt{x}}(x>0)\).
Solution:

Students often turn to ISC Class 12 Maths OP Malhotra Solutions Chapter 24 The Plane Ex 24(a) to clarify doubts and improve problem-solving skills.

S Chand Class 12 ICSE Maths Solutions Chapter 24 The Plane Ex 24(a)

Question 1.
Find the direction cosines of the normal to the plane
(i) 2x – 3y + 6z = 7
(ii) x + 2 y + 2 z – 1 = 0
Answer:
(i) Given eqn. of plane be 2 x – 3 y + 6z = 7
∴ direction ratios of normal to plane be <2, -3, 6>
Thus, direction cosines of normal to given plane be
< \(\frac{2}{\sqrt{2^2+(-3)^2+6^2}}\), \(\frac{-3}{\sqrt{2^2+(-3)^2+6^2}}\), \(\frac{6}{\sqrt{2^2+(-3)^2+7^2}}\) >
i.e.< \(\frac{2}{7}\), \(\frac{-3}{7}\), \(\frac{6}{7}\)>

(ii) Given eqn. of plane be
x + 2 y + 2 z – 1 = 0
∴ direction ratios of normal to plane be < 1, 2, 2 >
Thus, direction cosines of normal to plane are
< \(\frac{1}{\sqrt{1^2+2^2+2^2}}\), \(\frac{2}{\sqrt{1^2+2^2+2^2}}\), \(\frac{2}{\sqrt{1^2+2^2+2^2}}\) >
i.e.< \(\frac{1}{3}\), \(\frac{2}{3}\), \(\frac{2}{3}\)>

Question 2.
Show that the normals to the planes x – y + z = 1, 3 x + 2 y – z + 2 = 0 are inclined to each other at an angle of 90°.
Answer:
Given eqn. of plane are
x – y + 2 z = 1
and 3 x + 2 y – z + 2 = 0
∴ Direction numbers of normals to plane (1) and (2) are
< 1, -1, 1 > and < 3, 2, -1>
Here a₁ a₂ + b₁ b₂ + c₁ c₂
= 1(3) – 1(2) + 1(-1) = 0
Hence both normals are inclined to each other at an angle of 90°.

Question 3.
Find the intercepts of the plane 2 x – 3 y + 4 z = 12 on the coordinate axes.
Answer:
Given eqn. of plane be 2 x – 3 y + 4 z = 12
⇒ \(\frac{x}{6}\) – \(\frac{y}{4}\) + \(\frac{z}{3}\) = 1
Thus the intercepts made by plane (1) an coordinate axes are 6,-4 and 3 .

Question 4.
Write the equation of each of the following planes in the intercept form:
(i) 2 x – 3 y + 4 z = 12
(ii) 3 x + 2 y – 5 z = 15
(iii) x + 3 y + 4 z = 12
Answer:
(i) Given eqn. of plane be,
2 x – 3 y + 4 z = 12
On dividing throughout eqn. (1) by 12 , we have
\(\frac{2 x}{12}\) – \(\frac{3 y}{12}\) + \(\frac{4 z}{12}\) = 1
⇒ \(\frac{x}{6}\) – \(\frac{y}{4}\) + \(\frac{z}{3}\) = 1
which is the required intercept form of plane.

(ii) Given eqn. of plane be
3 x + 2 y – 5 z = 15
⇒ \(\frac{3 x}{15}\) + \(\frac{2 y}{15}\) – \(\frac{5 z}{15}\) = 1
⇒ \(\frac{x}{5}\) + \(\frac{y}{\frac{15}{2}}\) – \(\frac{z}{3}\) = 1
⇒ \(\frac{x}{5}\) + \(\frac{y}{\frac{15}{2}}\) + \(\frac{z}{-3}\) = 1
which is the required intercept from.

(iii) Given eqn. of plane be,
x + 3 y + 4 z = 12
⇒ \(\frac{x}{12}\) + \(\frac{y}{4}\) + \(\frac{z}{3}\) = 1
which is the required intercept form of plane

Question 5.
Find the equation of the plane
(i) which cuts the axes of x, y and z at (-2, 0, 0),(0, 3, 0),(0, 0, 5) respectively.
(ii) which cuts the axes of x and y at (3, 0, 0) and (0, -2, 0) and does not cut the z-axis.
(iii) which cuts the x-axis at (4, 0, 0) and does not cut the y-axis and the z-axis.
(iv) passing hrough the point (2, 3, 4) and making equal intercepts on the coordinates axes.
Answer:
(i) Let the eqn. of plane be
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
eqn. (1) passes through (-2, 0, 0)
∴ \(\frac{-2}{a}\) = 1
⇒ a = -2
eqn. (1) passes through (0, 3, 0) and (0, 0, 5)
∴ \(\frac{3}{b}\) = 1
⇒ b = 3
and \(\frac{5}{c}\) = 1
⇒ c = 5
Hence, the required eqn. of plane be
\(\frac{x}{-2}\) + \(\frac{y}{3}\) + \(\frac{z}{5}\) = 1

(ii) eqn. of plane parallel to z-axis be
\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
eqn. (1) passes through (3, 0, 0)
∴ \(\frac{3}{a}\) = 1
⇒ a = 3
Also, eqn. (1) passes through (0, -2, 0)
∴ \(-\frac{2}{b}\) = 1
⇒ b = -2
∴ eqn. (1) becomes,
\(\frac{x}{3}\) – \(\frac{y}{2}\) = 1
be the required eqn. of plane.

(iii) eqn. of plane does not cut y-axis and z-axis i.e. eqn. of plane parallel to y o z plane be
x = k
eqn. (1) given to be passes through (4, 0, 0)
∴ k = 4
Thus eqn. (1) becomes; x = 4 which gives the required eqn. of plane.

(iv) Let the eqn. of plane having equal intercepts a on coordinates axes be
\(\frac{x}{a}\) + \(\frac{y}{a}\) + \(\frac{z}{a}\) = 1
Also eqn. (1) passes through (2, 3, 4)
∴\(\frac{2}{a}\) + \(\frac{3}{a}\) + \(\frac{4}{a}\) = 1
⇒ a = 9
Thus eqn. (1) becomes:
x + y + z = 9
be the req. eqn. of plane

Question 6.
Find the equations of the two planes passing through the points (0, 4, -3) and (6, -4, 3), if the sum of their intercepts on the three axes is zero.
Answer:
Let the eqn. of plane be
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
It is given that sum of their intercepts an axes be 0
∴ a + b + c = 0
eqn. (1) passes through the point (0, 4, -3).
∴ \(\frac{0}{a}\) + \(\frac{4}{b}\) – \(\frac{3}{c}\) = 1
⇒ 4 c – 3 b = b c
Also the point (6, -4, 3) lies on eqn. (1); we have
\(\frac{6}{a}\) – \(\frac{4}{b}\) + \(\frac{3}{c}\) = 1
using eqn. (3) in eqn. (4); we have
\(\frac{6}{a}\) – 1 = 1
⇒ \(\frac{6}{a}\) = 2
⇒ a = 3
∴ from (2); c = -3 – b
∴ from (3); 4(-3 – b) – 3 b = b(- 3 – b)
⇒ – 12 – 7 b = -3 b – b²
⇒ b² – 4 b – 12 = 0
⇒ (b – 6)(b + 2) = 0
⇒ b = 6,-2
when b = 6
∴ c = -3 – b = -9
when b = -2
∴ c = -3 + 2 = -1
when a = 3 ; b = 6
and c = -9
∴ eqn. (1) becomes ; \(\frac{x}{3}\) + \(\frac{y}{6}\) + \(\frac{z}{-9}\) = 1
when a = 3 ; b = -2 and c = -1
∴ eqn. (1) becomes;
\(\frac{x}{3}\) + \(\frac{y}{-2}\) + \(\frac{z}{-1}\) = 1
be the reqd. eqns. of planes.

Question 7.
A plane meets the coordinate axes at A, B, C respectively such that the centroid of the triangle A B C is (1, -2, 3). Find the equation of the plane.
Answer:
Let the required eqn. of plane be
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
Then the coordinates of A, B and C are A (a, 0, 0), B(0, b, 0) and C(0, 0, c).
∴ centroid of the ∆ ABC are (\(\frac{a}{3}\), \(\frac{b}{3}\), \(\frac{c}{3}\)) also given centroid of ∆ ABC are (1, -2, 3)
i.e. \(\frac{a}{3}\) = 1
⇒ a = 3 ;
\(\frac{b}{3}\) = -2
⇒ b = -6
and \(\frac{c}{3}\) = 3
⇒ c = 9
putting the values of a, b, c in eqn. (1); we get
\(\frac{x}{3}\) + \(\frac{y}{-6}\) + \(\frac{z}{9}\) = 1
⇒ 6 x – 3 y + 2 z = 18
be the required equation of plane.

Question 8.
(i) Find the equation of the plane the normal to which from the origin is of length of 5 and which makes angles, with the axes of x, y and z, equal to 120°, 45° and 120° respectively.
(ii) Find the equation of the plane the normal to which from the origin is of length 4 and which makes angles with the axes of x, y, z equal to 90°, 135°, 45° respectively.
(iii) Find the equation of the plane such that the foot of the normal to which from the origin is the point (2, 3, 1).
(iv) Find the equation of the plane such that the length of the perpendicular to which from the origin is equal to 2 units, and the angles α, β, γ that this perpendicular makes with the axes of x, y, z are connected by the relation
\(\frac{\cos \alpha}{-1}\) = \(\frac{\cos \beta}{4}\) = \(\frac{\cos \gamma}{4}\) .
Answer:
(i) Here p = length of the normal from origin = 5
∴D cosines of normal to plane are < cos 120°, cos 45°, cos 120° >
i.e. < \(-\frac{1}{2}\), \(\frac{1}{\sqrt{2}}\), \(-\frac{1}{2}\) >
i.e. l = \(-\frac{1}{2}\) ; m = \(\frac{1}{\sqrt{2}}\) ; n = \(-\frac{1}{2}\)
Thus required eqn. of plane be
b x + m y + n z = p
\(\frac{-x}{2}\) + \(\frac{y}{\sqrt{2}}\) y – \(\frac{1}{2}\) z = 5
⇒ x – \(\sqrt{2}\) y + z + 10 = 0

(ii) Here p = length of the normal from origin = 4
∴ D cosines of normal to plane are < cos 90°, cos 135°, cos 45° >
i.e. < 0, \(-\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\) >
i.e. l = 0 ; m = \(-\frac{1}{\sqrt{2}}\) ; n = \(\frac{1}{\sqrt{2}}\)
Hence the required eqn. of plane be
b e + m y + n z = p
⇒ 0 x – \(\frac{1}{\sqrt{2}}\) y + \(\frac{1}{\sqrt{2}}\) z = 4
⇒ -y + z = 4 \(\sqrt{2}\)
⇒ y – z + 4 \(\sqrt{2}\) = 0
be the required eqn. of plane

(iii) Let the eqn. of plane through the point (2, 3, 1) be given by
a(x – 2) + b(y – 3) + c(z – 1) = 0 .

Where < a, b, c > are the direction ratios of the normal to plane
∴ direction ratios of normal to plane ON are < 2, 3, 1 >
Thus eqn. (1) becomes;
2(x – 2) + 3(y – 3) + 1(z – 1) = 0
⇒ 2 x + 3 y + z – 14 = 0
be the reqd. eqn. of plane.

(iv) Given p = length of ⊥ to the reqd. plane from origin =2 units
Since α, β , and γ are the angles made by this perpendicular with coordinates areas
Then direction cosines of this normal are < cosα, cosβ , cosγ >
Given \(\frac{\cos \alpha}{-1}\) = \(\frac{\cos \beta}{4}\) = \(\frac{\cos \gamma}{8}\)
= \(\frac{\sqrt{\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma}}{\sqrt{1+16+64}}\)
= ± \(\frac{1}{9}\)
⇒ cos α = ± \(\frac{1}{9}\) ; cos β = ± \(\frac{4}{9}\)
and cos γ = ± \(\frac{8}{9}\)
Thus direction cosines of normal to plane are < ± \(\frac{1}{9}\), ± \(\frac{4}{9}\), ± \(\frac{8}{9}\) >
i.e. l = ± \(\frac{1}{9}\) ; m = ± \(\frac{4}{9}\) and n = ± \(\frac{8}{9}\)
Hence the reqd. eqn. of plane be
b e + m y + n z = p
\(\frac{-1}{9}\) x + \(\frac{4}{9}\) y + \(\frac{8}{9}\) z = 2
i.e x – 4 y – 8 z + 18 = 0
and \(\frac{x}{9}\) – \(\frac{4 y}{9}\) – \(\frac{8 z}{9}\) = 2
⇒ x – 4 y – 8 z – 18 = 0

Question 9.
Reduce each of the following equations to the normal form and then determine the direction cosines and the length of the normal from the origin :
(i) 2 x – 2 y + z – 12 = 0
(ii) 9 x + 6 y – 2 z + 7 = 0
Answer:
(i) Given eqn. of plane be
2 x – 2 y + z – 12 = 0
Dividing eqn. (1) throughout by \(\sqrt{2^2+(-2)^2+1^2}\) i.e. 3 , we get
\(\frac{2}{3}\) x – \(\frac{2}{3}\) y + \(\frac{z}{3}\) – \(\frac{12}{3}\) = 0
\(\frac{2}{3}\) x – \(\frac{2}{3}\) y + \(\frac{z}{3}\) = 4
which is the required normal form of plane.
On comparing with lx + m y + n z = p
Here l = \(\frac{2}{3}\) ; m = \(-\frac{2}{3}\)
and n = \(\frac{1}{3}\) and p = 4
Thus, the direction cosines of normal to plane are < \(\frac{2}{3}\), \(-\frac{2}{3}\), \(\frac{1}{3}\) >
and p = length of normal from origin = 4.

(ii) Given eqn. of plane be
9 x + 6 y – 2 z + 7 = 0
on dividing eqn. (1) throughout by
\(\sqrt{9^2+6^2+c-2^2}\)
i.e. \(\sqrt{81+36+4}\)
i.e. 11 ; we have
\(\frac{9}{11}\) x + \(\frac{6}{11}\) y – \(\frac{2}{11}\) z + \(\frac{7}{11}\) = 0
⇒ \(\frac{9 x}{11}\) + \(\frac{6 y}{11}\) – \(\frac{2 z}{11}\) = \(\frac{-7}{11}\)
⇒ \(\frac{-9}{11}\) x – \(\frac{6}{11}\) y + \(\frac{2}{11}\) z = \(\frac{7}{11}\)
which is the required normal form and on comparing with l x + m y + n z = p
Here, l = \(-\frac{9}{11}\) ; m = \(-\frac{6}{11}\)
and n = \(\frac{z}{11}\) and p = \(\frac{7}{11}\).
Thus reqd. direction cosines of normal to plane are < \(\frac{-9}{11}\), \(\frac{-6}{11}\), \(\frac{2}{11}\) >
and p = length of normal from origin = \(\frac{7}{11}\)

Question 10.
For each of the following planes, find the direction ratios and the direction cosines of the normal from the origin.
(i) 3 x + 2 y – 6 z + 22 = 0
(ii) 4 x – 3 y + 5 z – 30 = 0
(iii) 5 y – 12 z + 20 = 0
Answer:
(i) Given eqn. of plane be
3 x+2 y-6 z+2 z=0
∴ Direction ratios of normal to plane are < 3, 2, -6 >
⇒ 3 x + 2 y – 6 z = -22
⇒ -3x – 2 y + 6 z = 22
on dividing eqn. (1) by
(-2)² + 6² i.e. 7 ;
we get
\(\frac{-3}{7}\) x – \(\frac{2}{7}\) y + \(\frac{6}{7}\) z = \(\frac{22}{7}\)
Thus the required direction cosines of normal to plane are < \(\frac{-3}{7}\), \(\frac{-2}{7}\), \(\frac{6}{7}\)

(ii) Given eqn. of plane be
4 x – 3 y + 5 z – 30 = 0
∴ direction ratios of normal to plane are < 4,-3,5 >
⇒ 4 x – 3 y + 5 z = 30
on diviaing throughout by \(\sqrt{16+9+25}\) i.e. 5 \(\sqrt{2}\); we get
\(\frac{4}{5 \sqrt{2}}\) x – \(\frac{3}{5 \sqrt{2}}\) y + \(\frac{1}{\sqrt{2}}\) z = \(\frac{30}{5 \sqrt{2}}\)
on comparing with b x + m y + n z = p
Here l = \(\frac{4}{5 \sqrt{2}}\) ; m = \(\frac{-3}{5 \sqrt{2}}\) and n = \(\frac{1}{\sqrt{2}}\) Hence the direction cosines of normal to plane are < \(\frac{4}{5 \sqrt{2}}\), \(\frac{-3}{5 \sqrt{2}}\), \(\frac{1}{\sqrt{2}}\) >.

(iii) Given eqn. of plane be 5 y – 12z = -20
∴ direction ratios of normal to plane are < 0, 5, -12 >
∴ -5 y + 12 z = 20
on dividing eqn. (1) throughout by
\(\sqrt{(-5)^2+12^2}\) = \(\sqrt{169}\) = 13 ; we get
Thus, the direction cosines of normal to plane are < 0, \(\frac{-5}{13}\), \(\frac{12}{13}\)

Question 11.
(i) Find the coordinates of the point, where the line \(\frac{x+1}{2}\) = \(\frac{y+2}{3}\) = \(\frac{z+3}{4}\) meets the plane x + y + 4 z = 6.
(ii) Find the coordinates of the point where the line joining the points (1, -2, 3) and (2, -1, 5) cuts the plane x – 2 y + 3 z = 1. Hence, find the distance of this point from the point (5, 4, 1).
(iii) Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the X Z plane: Also find the angle which the line makes with the X Z plane.
Answer:
(i) Given eqn. of line be
\(\frac{x+1}{2}\) = \(\frac{y+2}{3}\) = \(\frac{z+3}{4}\) = t (say)
So any point on line (1) be P(2 t – 1, 3 t – 2, 4 t – 3)
Given eqn. of plane be
x + y + 4 z = 6
For the point of intersection of line (1) and plane (2)
∴ point P lies on eqn. (2).
∴ 2 t – 1 + 3 t – 2 + 4(4 t – 3) = 6
⇒ 21 t = 21
⇒ t = 1
Hence the required point of intersection be P(2 – 1,3 – 2,4 – 3) i.e. P(1, 1, 1).

(ii) eqn. of line through the points (1, -2, 3) and (2, -1, 5) is given by
\(\frac{x-1}{2-1}\) = \(\frac{y+2}{-1+2}\) = \(\frac{z-3}{5-3}\)
i.e. \(\frac{x-1}{1}\) = \(\frac{y+2}{1}\) = \(\frac{z-3}{2}\)
∴ any point on line (1) be given by
\(\frac{z+1}{1}\) = \(\frac{y+2}{1}\) = \(\frac{z-3}{2}\) = t (say)
P (t + 1, t – 2, 2 t + 3)
Also given eqn. of plane be
x – 2 y + 3 z = 19
For point of intersection of line (1) and plane (2), the point P(t + 1, t – 2,2 t + 3) lies on plane (2).
∴ t + 1 – 2(t – 2) + 3(2 t + 3) = 19
⇒ 5 t + 14 = 19 ⇒ 5 t = 5 ⇒ t = 1
Thus the required point of intersection of given line (1) and (2) be
P(1 + 1, 1 – 2, 2 + 3) i.e. P(2, -1, 5)
∴ required distance of P(2, -1, 5) from given point (5, 4, 1)
= (5 – 2)² + (4 + 1)² + (1 – 5)²
= \(\sqrt{9+25+16}\) = \(\sqrt{50}\) = 5 \(\sqrt{2}\) units

(iii) eqn. of line passing through the points A(3, 4, 1) and B(5, 1, 6) is given by
\(\frac{x-3}{5-3} \) = \(\frac{y-4}{1-4}\) = \(\frac{z-1}{6-1}\)
i.e. \(\frac{x-3}{2}\) = \(\frac{y-4}{-3}\) = \(\frac{z-1}{5}\)
Any point on line (1) be
i.e \(\frac{x-3}{2}\) = \(\frac{y-4}{-3}\) = \(\frac{z-1}{5}\) = t (say)
given by P(2 t + 3, -3 t + 4,5 t + 1)
eqn. of XOZ plane be y=0
Since the line (1) crosses XOZ plane
∴ point P lies on eqn. (2)
∴-3 t + 4 = 0 ⇒ t = 4 / 3
∴ required point of intersection of line (1) and plane ( 2 ) be:
P (\(\frac{8}{3}\) + 3, -4 + 4, \(\frac{20}{3}\) + 1)
i.e. P (\(\frac{17}{3}\), 0, \(\frac{23}{3}\))
Let θ be the angle between plane (2) and line (1)
Then 90° – θ be the angle between line (1) and normal to plane (2).
Now, direction ratios of line (1) are < 2, -3, 5 >and direction ratios of normal to plane (2) are < 0, 1, 0 >.
∴ cos (90° – θ) = \(\frac{|2(0)-3(1)+5(10)|}{\sqrt{2^2+9+25} \sqrt{0^2+1^2+0}}\)
= \(\frac{3}{\sqrt{38}}\)
⇒ sinθ
= \(\frac{3}{\sqrt{38}}\)

Question 12.
Foot of the perpendicular from the origin to the plane is (2, 3, 4). Find the equation of the Plane.
Answer:
The required eqn. of plane through the point N(2, 3, 4) be given by
a(x – 2) + b(y – 3) + c(z – 4) = 0

Where < a, b, c >are the direction numbers of normal to piane.
Further ON be the normal to the plane (1) Also, direction ratios of ON are < 2, 3, 4 >
∴\(\frac{a}{2}\) = \(\frac{b}{3}\) = \(\frac{c}{4}\) = k (say)
i.e. a = 2 k ; b = 3 k and c = 4 k
∴ eqn. (1) becomes ;
2(x – 2) + 3(y – 3) + 4(z – 4) = 0
⇒ 2 x + 3 y + 4 z – 29 = 0
be the reqd. eqn. of plane.

Question 13.
Find the distance of the point (2, 3, 4) from the plane 3 x + 2 y + 2 z + 5 = 0, measured parallel to the line
\(\frac{x+3}{3}\) = \(\frac{y-2}{6}\) = \(\frac{z}{2}\) .
Answer:
eqn. of given line be
\(\frac{x+3}{3}\) = \(\frac{y-2}{6}\) = \(\frac{z}{2}\)
eqn.of any line through point A(2, 3, 4) and |l to line (1) be given by
\(\frac{x-2}{3}\) = \(\frac{y-3}{6}\) = \(\frac{z-4}{2}\)

So any point an line (2) be given by
\(\frac{x-2}{3}\) = \(\frac{y-3}{6}\) = \(\frac{z-4}{2}\) = t (say)
P(3 t + 2, 6 t + 3, 2 t + 4)
Since point P lies on given plane
3 x + 2 y + 2 z + 5 = 0
i.c. 3(3 t + 2) + 2(6 t + 3) + 2(2 t + 4) + 5 = 0
⇒ 25 t + 25 = 0 ⇒ t = -1
Hence the coordinates of P becomes
(-3 + 2, -6 + 3, -2 + 4) i.e. P(-1, -3, 2)
Thus required distance = |AP|
= \(\sqrt{(-1-2)^2+(-3-3)^2+(2-4)^2}\)
= \(\sqrt{9+36+4}\) = 7 units

Question 14.
Find the equation in Cartesian form of the plane passing through the point (3, – 3, 1) and normal to the line joining the points (3, 4, -1) and (2, -1, 5).
Answer:
The eqn. of plane through the point (3, -3, 1) be given by
a(x – 3) + b(y + 3) + c(z – 1) = 0
Where < a, b, c >are the direction ratios of normal to plane.
Also direction ratios of normal to plane are
< 2 – 3, -1 – 4, 5 + 1 > i.e. < -1, -5, 6 >
∴ \(\frac{a}{-1}\) = \(\frac{b}{-5}\) = \(\frac{c}{6}\) = k (say)
a = -k ; b = -5 k ; c = 6 k
putting all these values in eqn. (1); we have
⇒ -k(x – 3) – 5 k(y + 3) + 6 k(z – 1) = 0
⇒ -x – 5 y + 6 z – 18 = 0
⇒ x + 5 y – 6 z + 18 = 0
be the required eqn. of plane.

Question 15.
Find the image of the point
(i) (3, -2, 1) in the plane
3 x – y + 4 z = 2
(ii) (0,0,0) in the plane
3 x + 4 y – 6 z + 1 = 0
Answer:
(i) Let M be the foot of ⊥ drawn from P(3, -2, 1) on given plane
3 x – y + 4 z = 2
∴ eqn. of line PM i.e. line passing through the point P(3, -2, 1) and normal to plane (1) be
\(\frac{x-3}{3}\) = \(\frac{y+2}{-1}\) = \(\frac{z-1}{4}\) = t (say)

So any point on line (2) be
Q(3 t + 3, -t – 2, 4 t + 1)
NowQ be the image of P in plane (1) if mid point of PQ i.e. M lies on plane (1) Here coordinates of
M (\(\frac{3 t+3+3}{2}\), \(\frac{-t-2-2}{2}\), \(\frac{4 t+1+1}{2}\))
i.e. M (\(\frac{3 t+6}{2}\), \(\frac{-t-4}{2}\), \(\frac{4 t+2}{2}\))
Now M lies on plane (1)
∴ 3(\(\frac{3 t+6}{2}\)) – (\(\frac{-t-4}{2}\)) + 4(\(\frac{4 t+2}{2}\)) = 2
⇒ 9 t + 18 + t + 4 + 16 t + 8 = 4
⇒ 26 t + 26 = 0
⇒ t = -1
Hence the required coordinates of image of given point in given plane be
Q (3 – 3, + 1 – 2, -4 + 1)
i.e. Q(0, -1, -3)

(ii) Let M be the foot of ⊥ drawn from O(0, 0, 0) on given plane
3 x + 4 y – 6 z + 1 = 0
This eqn. of line OM i.e. line passing through the point O(0, 0, 0) and norinal to plane (1) be
\(\frac{x-0}{3}\) = \(\frac{y-0}{4}\) = \(\frac{z-0}{-6}\) = t (say)

So any point on line (2) be P(3 t, 4 t, – 6 t)
Now P be the image of point O in given plane
Then mid point of OP i.e. M lies on given plane (1)
Here coordinates of M(\(\frac{3 t}{2}\), \(\frac{4 t}{2}\), \(\frac{-6 t}{2}\)) and lies on plane (1)
3(\(\frac{3 t}{2}\)) + 4(\(\frac{4 t}{2}\)) – 6(\(-\frac{6 t}{2}\)) + 1 = 0
⇒ 9 t + 16 t + 36 t + 2 = 0
⇒ t = \(-\frac{2}{61}\)
Hence required coordinates of image of given point in plane (1) be
P( \(-\frac{6}{61}\), \(\frac{-8}{61}\), \(\frac{12}{61}\)).

Question 16.
Find the reflection of the point (1, 2, -1) in the plane 3 x – 5 y + 4 z = 5.
Answer:
Let M be the foot of ⊥ drawn from given point P(1, 2, -1) on given plane
3 x – 5 y + 4 z = 5
Thus eqn. of PM i.e. the line through the point P(1, 2, -1) and normal to plane (1) be given by
\(\frac{x-1}{3}\) = \(\frac{y-2}{-5}\) = \(\frac{z+1}{4}\) = t (say)

So any point on line (2) be
Q(3 t + 1, -5 t + 2, 4 t – 1)
Now point Q be the image of P in plane (1) if the mid point of line segment P Q
i.e. M (\(\frac{3 t+1+2}{2}\), \(\frac{-5 t+2+2}{2}\), \(\frac{4 t-1-1}{2}\))
lies on plane (1)
∴ 3(\(\frac{3 t+2}{2}\)) – 5(\(\frac{-5 t+4}{2}\)) + 4(\(\frac{4 t-2}{2}\)) = 5
⇒ 9 t + 6 + 25 t – 20 + 16 t – 8 = 10
⇒ 50 t = 32 ⇒ t = \(\frac{16}{25}\)
Hence the required coordinates of reflection of point P be Q(\(\frac{48}{25}\) + 1, \(\frac{-80}{25}\) + 2, \(\frac{64}{25}\) – 1) i.e. Q(\(\frac{73}{25}\), \(\frac{-6}{5}\), \(\frac{39}{25}\))

Question 17.
From the point P(1, 2, 4), a perpendicular is drawn on the plane 2 x + y – 2 z + 3 = 0. Find the equation, the length and the coordinates of the foot of the perpendicular.
Answer:
Let M be the foot of ⊥ drawn from given point P(1, 2, 4) on given plane
2 x + y – 2 z + 3 = 0
∴ eqn. of line PM i.e. the line through P(1, 2, 4) and normal to plane (1) be given by
\(\frac{x-1}{2}\) = \(\frac{y-2}{1}\) = \(\frac{z-4}{-2}\) = t (say)
So any point on line (2) be M(2 t + 1, t + 2, -2 t + 4) clearly M lies on plane (1).
∴ 2(2 t + 1) + t + 2 – 2(-2 t + 4) + 3 = 0

⇒ 4 t + 2 + t + 2 + 4 t – 8 + 3 = 0
⇒ 9 t – 1 = 0 ⇒ t = \(\frac{1}{9}\)
∴ Coordinates of foot of ⊥ M be given by
(\(\frac{2}{9}\) + 1, \(\frac{1}{9}\) + 2, \(\frac{-2}{9}\) + 4)
i.e. M(\(\frac{11}{9}\), \(\frac{19}{9}\), \(\frac{34}{9}\))
and
|P M| = \(\sqrt{(\frac{11}{9}-1)^2 + (\frac{19}{9}-2)^2 + (\frac{34}{9}-4)^2)}\)
= \(\sqrt{\frac{4}{81}+\frac{1}{81}+\frac{4}{81}}\) = \(\sqrt{\frac{1}{9}}\) = \(\frac{1}{3}\) units

Question 18.
Find the distance of the point (1, -2, 3) from the plane x – y + z = 5 measured along a line parallel to \(\frac{x}{2}\) = \(\frac{y}{3}\) = \(\frac{z}{-6}\).
Answer:
Given eqn. of line be
\(\frac{x}{2}\) = \(\frac{y}{3}\) = \(\frac{z}{-6}\)
Thus eqn. of line through the given point A(1, -2, 3) and parallel to line (1) be given by
\(\frac{x-1}{2}\) = \(\frac{y+2}{3}\) = \(\frac{z-3}{-6}\) = t (say)
So any point on line (2) be
P(2 t + 1, 3 t – 2, -6 t + 3)
Since the point P} lies on given plane
x – y + z = 5
2 t + 1 – (3 t – 2) + (-6 t + 3) = 5
⇒ -7 t = -1
⇒ t = 1 / 7

Thus required coordinates of point P are (\(\frac{2}{7}\) + 1, \(\frac{3}{7}\) – 2, \(\frac{-6}{7}\) + 3)
i.e. P (\(\frac{9}{7}\), \(\frac{-11}{7}\), \(\frac{15}{7}\))
Thus required distance = |A D|
= \(\sqrt{(\frac{9}{7}-1)^2 + (\frac{-11}{7}+2)^2 + (\frac{15}{7}-3)^2}\)
= \(\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}\)
= \(\sqrt{\frac{49}{49}}\) = 1 units.

Question 19.
Prove that a variable plane which moves so that the sum of reciprocals of its intercepts on the three coordinate axes is constant, passes through a fixed point.
Answer:
Let the eqn. of variable plane be
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
Where a, b, c are the lengths of intercepts cut off by plane on coordinates ares.
also it is given that
\(\frac{1}{a}\) + \(\frac{1}{b}\) + \(\frac{1}{c}\) = k (constant)
⇒ \(\frac{1}{k a}\) + \(\frac{1}{k b}\) + \(\frac{1}{k c}\) = 1
⇒ \(\frac{1}{a}\) (\(\frac{1}{k}\)) + \(\frac{1}{b}\) (\(\frac{1}{k}\)) + \(\frac{1}{c}\) (\(\frac{1}{k}\)) =1
eqn. (2) shows that eqn. (1) passes through (\(\frac{1}{k}\), \(\frac{1}{k}\), \(\frac{1}{k}\)) which is a fixed point, since k be a constant.
Hence the variable plane passes through fixed point.

Question 20.
(i) A variable plane is at a constant distance p from the origin and meets the axes in A, B, C. Show that the locus of the centroid of triangle A B C is
x^-2 + y^-2 + z^-2 = 9 p^-2
(ii) A variable plane, which remains at a constant distance of 9 units from the origin, cuts the coordinates axes at the points A, B and C. Show that the locus of the centroid of ∆ MBC is
\(\frac{1}{x^2}\) + \(\frac{1}{y^2}\) + \(\frac{1}{z^2}\) = \(\frac{1}{9}\)
Answer:
(i) Let the eqn. of variable plane be
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
p = ⊥ distance from (0, 0, 0) an plane (1)
⇒ p = \(\frac{\left|\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1\right|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\)
⇒ \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\) = \(\frac{1}{p}\) ;
on squaring ; we have
\(\frac{1}{a^2}\) + \(\frac{1}{b^2}\) + \(\frac{1}{c^2}\) = \(\frac{1}{p^2}\)
Since the plane (1) meets x-axis (y = 0 = z) at A
i.e: \(\frac{x}{a}\) = 1 ⇒ x = a
∴ coordinates of A are (a, 0,0)
Now plane (1) meets y-axis (x = 0 = z) at B
∴ from (1); \(\frac{y}{b}\) = 1 ⇒ y = b
Thus coordinates of point B are (0, b, 0) Hence (1) meets z-axis (x = 0 = y) at point C
∴ from (1); \(\frac{z}{c}\) = 1 ⇒ z = c
∴ coordinates of point C are (0,0, c) Thus centroid of ∆ABC be given by
(\(\frac{a+0+0}{3}\), \(\frac{0+b+0}{3}\), \(\frac{0+0+c}{3}\))
i.e. (\(\frac{a}{3}\), \(\frac{b}{3}\), \(\frac{c}{3}\))
Also, let the centroid of ∆ABC be G(α, β, γ)
∴ α = \(\frac{a}{3}\), β = \(\frac{b}{3}\) and γ = \(\frac{c}{3}\)
⇒ a = 3 α, b = 3 β and c = 3 γ
Putting the values of a, b and c in eqn. (1); we get
\(\frac{1}{9 α^2}\) + \(\frac{1}{9 β^2}\) + \(\frac{1}{9 γ^2}\) = \(\frac{1}{p^2}\)
Hence the locus of G(α, β, γ) be given by
[i.e. replacing α by x ; β by y and γ by z ]
∴ from (2); \(\frac{1}{9 x^2}\) + \(\frac{1}{9 b^2}\) + \(\frac{1}{9 z^2}\) = \(\frac{1}{p^2}\)
∴ x^-2 + y^-2 + z^-2 = 9 p^-2

(ii) Let the eqn. of variable plane be
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
given 9 = ⊥ distance from (0, 0, 0) on plane (1)
⇒ 9 = \(\frac{|0+0+0-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\)
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\) = \(\frac{1}{9}\)
on squaring; we have
\(\frac{1}{a^2}\) + \(\frac{1}{b^2}\) + \(\frac{1}{c^2}\) = \(\frac{1}{81}\)
eqn. (1) m eets x-axis at A i.e. y = 0 = z
∴ from (1); \(\frac{x}{a}\) = 1 ⇒ x = a
∴ coordinates of A} are (a, 0,0) eqn. (1) meets y-axis at B i.e. x = 0 = z
∴ from (1); \(\frac{y}{b}\) = 1 ⇒ y = b
∴ coordinates of B} are (0, b, 0)
eqn. (1) meets z-axis at c i.e. x = 0 = y
∴ from (1); \(\frac{z}{c}\) = 1 ⇒ z = c
∴ coordinates of C are (0,0, c)
∴ centroid of ∆ABC be (\(\frac{a}{3}\), \(\frac{b}{3}\), \(\frac{c}{3}\))
Also but centroid of ∆ABC be G(α, β, γ)
i.e. \(\frac{a}{3}\) = α ⇒ a = 3 α
and \(\frac{c}{3}\) = γ ⇒ c = 3 γ
and \(\frac{b}{3}\) = β ⇒ b = 3 β
∴ From (2); \(\frac{1}{α^2}\) + \(\frac{1}{β^2}\) + \(\frac{1}{γ^2}\) = \(\frac{9}{81}\) = \(\frac{1}{9}\)
Thus, the locus of G(α, β, γ) be given by
\(\frac{1}{x^2}\) + \(\frac{1}{y^2}\) + \(\frac{1}{z^2}\) = \(\frac{1}{9}\)
[Replacing α by x ; β by y and γ by z ].

Question 21.
Find the equation of the plane passing through the following points :
(i) A(2, 2, -1), B(3, 4, 2), Q(7, 0, 6)
(ii) A(2, 1, 0), B(3, -2, -2), C(3, 1, 7).
Answer:
(i) Let the eqn. of plane through the point A (2, 2, -1) be given by a(x – 2) + b(y – 2) + c(z + 1) = 0 ……….. (1)
Where < a, b, c > be the direction ratios of normal to plane (1)
plane (1) passes through the point B(3, 4, 2)
∴ a(3 – 2) + b(4 – 2) + c(2 + 1) = 0
i.e. a + 2 b + 3 c = 0
Also point C(7, 0, 6) lies on plane
a(7 – 2) + b(0 – 2) + c(6 + 1) = 0
5 a – 2 b + 7 c = 0
On solving eqn. (2) and (3) by using cross-multiplication methods we have
\(\frac{a}{14+6}\) = \(\frac{b}{15-7}\) = \(\frac{c}{-2-10}\)
⇒ \(\frac{a}{20}\) = \(\frac{b}{8}\) = \(\frac{c}{-12}\) = k (say) ; k < 0
⇒ a = 20 k ; b = 8 k ; c = -12 k
Putting the values of a, b & c in eqn. (1); we have
20 k(x – 1) + 8 k(y – 2) – 12 k(z + 1) = 0
⇒ 20 x + 8 y – 12 z – 48 = 0
⇒ 5 x + 2 y – 3 z – 12 = 0 .

(ii) eqn. of any plane through te point (2, 1, 0) is given by
a(x – 2) + b(y – 1) + c(z – 0) = 0
Since the plane passes through given points (3, -2, -2) and (3, 1, 7). Thus (3, -2, -2) and (3, 1, 7) lies on eqn. (1). a(3 – 2) + b(-2 – 1) + c(-2 – 0) = 0
⇒ a – 3 b – 2 c = 0
also, a(3 – 2) + b(1 – 1) + c(7 – 0) = 0
⇒ a + 0 b + 7 c = 0
On solving (2) and (3) ; we have
\(\frac{a}{-21-0}\) = \(\frac{b}{-2-7}\) = \(\frac{c}{0+3}\)
i.e. \(\frac{a}{-21}\) = \(\frac{b}{-9}\) = \(\frac{c}{3}\) = k (say)
i.e. a = -21 k ; b = -9 k ; c = 3 k
∴ From eqn. (1); we have
-21 k(x – 2) – 9 k(y – 1) + 3 k(z – 0) = 0
⇒ -3 k [7 x – 14 + 3 y – 3 – z] = 0
⇒ 7 x + 3 y – z = 17
be the required equation of plane.

Question 22.
Show that the following points are coplanar: (-6, 3, 2), (3, -2, 4), (5, 7, 3) and (-13, 17, -1)
Answer:
For proving any four points A, B, C D are coplanar if plane through any three points must pass through the remaining fourth point.
Let the eqn.of plane through the point A(-6, 3, 2) is given by
a(x + 6) + b(y – 3) + c(z – 2) = 0
Now plane (1) passes through the point B (3, -2, 4)
a(3 + 6) + b(-2 – 3) + c(4 – 2) = 0
∴ 9 a – 5 b + 2 c = 0
Again the point C (5, 7, 3) lies on plane (1).
∴ a(5 + 6) + b(7 – 3) + c(3 – 2) = 0
i.e. 11 a + 4 b + c = 0
On solving eqn. (2) and eqn. (3) using crossmultiplication method, we have
\(\frac{a}{-5-8}\)
= \(\frac{b}{22-9}\)
= \(\frac{c}{36+55}\)
i.e. \(\frac{a}{-13}\)
= \(\frac{b}{13}\)
= \(\frac{c}{91}\) = k (say) ; k < 0
⇒ a = -13 k ; b = 13 k ; c = 91
k putting the values of a, b c in eqn. (1); we have
-13 k(x + 6) + 13 k(y – 3) + 91 k(z – 2) = 0
⇒ -13 x + 13 y + 91 z – 299 = 0
⇒ x – y – 7 z + 23 = 0
Further point D lies on eqn …………… (4)
∴ D(-13, 17, -1) satisfies eqn …………… (4)
i.e. -13 – 17 + 7 + 23 = 0
⇒ 0 = 0, which is true
Hence all the given four points lies in a same plane
∴ all the given points are coplanar.

Question 23.
Show that the line joining the points (0, -1, 2) and (2, -1, -1) is coplanar with the line joining the points (1, 1, 1) and (0, 3, 3).
Answer:
Eqn. of plane through the point (0, -1, 2) is given by
a(x – 0) + b(y + 1) + c(z – 2) = 0
Now the point (2, -1, -1) lies on plane (1).
∴ a(2 – 0) + b(-1, + 1) + c(-1 – 2) = 0
⇒ 2 a + 0 b – 3 c = 0
Again the point (1, 1, 1) lies on plane (1).
∴ a + 2 b – c = 0
Solving eqn. (2) (3) by using crossmultipluation method, we have
\(\frac{a}{0+6}\)
= \(\frac{b}{-3+2}\) + \(\frac{c}{4-0}\)
\(\frac{a}{6}\)
= \(\frac{b}{-1}\) = \(\frac{c}{4}\) = k (say)
∴ a = 6 k ; b = -k ; c = 4 k where k < 0
Putting all these values in eqn. (1); we have
6 k x – k(y + 1) + 4 k(z – 2) = 0
⇒ 6 x – y + 4 z – 9 = 0
Now the remaining point (0, 3, 3) lies on plane (4) if (0, 3, 3) satisfies eqn. ………… (4).
if 6 × 0 – 3 + 12 – 9 = 0 if 0 = 0, which is true.
Hence all the four points (0, -1, 2), (2, -1, -1), (1, 1, 1) and (0, 3, 3).

Utilizing ISC S Chand Maths Class 12 Solutions Chapter 7 Continuity and Differentiability of Functions Ex 7 as a study aid can enhance exam preparation.

S Chand Class 12 ICSE Maths Solutions Chapter 7 Continuity and Differentiability of Functions Ex 7

Question 1.
Examine the continuity of the function f(x) = 2x² – 1 at x = 3.
Solution:
Given f(x) = 2x² – 1

Thus f(x) is continous at x = 3

Question 2.
Examine the following functions for continuity:
(a) f(x) = x- 5
(b) f(x) = \(\frac{1}{x-5}\), x ≠ 5
(c) f(x) = \(\frac{x^2-25}{x+5}\), x ≠ 5
(d) f(x) = |x – 5|
Solution:
(a) Given f (x) = x – 5 ; Df = R
Let c ∈ Dy = be any arbitrary point.
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x – 5 = c – 5 = f(c)
∴ f is continuous at x = c
but c be any arbitrary point of D_f
Thus,/is continuous at every point of its domain
∴ f be a continuous function.

(b) f(x) = \(\frac{1}{x-5}\), x ≠ 5
Here, D_f = R -{5}
Let c ∈ D_f = be any arbitrary point.

Thus, f is continuous at x = c
and c be any arbitrary point of D_f
Therefore f is continuous at every point of its domain

(c) Given \(\frac{x^2-25}{x+5}\), x ≠ 5
Here, D_f = R -{5}
Let c ∈ D_f be any arbitrary point.
∴ c ≠ – 5
Then \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) \(\frac{x^2-25}{x-5}=\frac{c^2-25}{c-5}\)
since c ≠ – 5
∴ f(c) = \(\frac{c^2-25}{c-5}\)
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\)f(x) = f(c)
Thus, f is continuous at x = c
but c be any arbitrary point
Hence f is continuous at every point of its domain.

(d) Given f(x) = |x – 5|
= \(\left\{\begin{array}{cc}
x-5 ; & x \geq 5 \\
-(x-5) ; & x<5
\end{array}\right.\)
D_f = R
So we examine the continuity of function of f at all x ∈ R. Let c ∈ R be any arbitrary point of D_f
There cases arises.
Case -I :
when c < 5
Then f(c) = -(c – 5)
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 9 (x – 5) = – (c – 5)
Thus, \(\underset{x \rightarrow c}{\mathrm{Lt}}\)f(x) = f(c)
∴ f(x) is continous for all c < 5

Case -II : when c > 5
Then f(c) = c – 5
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x – 5 = c – 5
Thus, \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = f(c)
∴ f is continuous for all c > 5

Case -III :
when x = C = 5
Then f(c) = f(5) = 5 – 5 = 0

Thus f(x) is continous at x = c = 5
on combining all three cases; function f is continuous for all x ∈ R.
Thus f is continuous at every point of its domain.

Question 3.
A function f is defined by
f(x) = \(\left.\begin{array}{ll}
=\frac{x^2-4 x+3}{x^2-1}, & \text { for } x \neq 1 \\
=2, & \text { for } x=1
\end{array}\right\}\)
Test the continuity of the function at x = 1.
Solution:
Given

also f(1) = 2
∴ L.H.L = R.H.L = f(1)
Thus f is di continous at x = 1

Question 4.
Prove that the funciotn f(x) = xⁿ is continuous at x = n, where n is a positive integer.
Solution:

Question 5.
Show that the function
f(x) = \(\left\{\begin{array}{c}
x^2 \text { for } 1 \leq x<2 \\
3 x-4 \text { for } 2 \leq x<4
\end{array}\right.\)
is discontinuous at x = 2 and continuous x = 3.
Solution:

Question 6.
(a) Find all points of discontinuity of f where f is defined by
f(x) = \(\left\{\begin{array}{l}
2 x+3, \text { if } x \leq 2 \\
2 x-3, \text { if } x>2
\end{array}\right.\)
(b) Discuss the discontinuity of the function
f(x) at x = 0 , if f(x) = \(\left\{\begin{array}{l}
2 x-1, \text { if } x<0 \\
2 x+1, \text { if } x \geq 0
\end{array}\right.\)
(c) Is the function defined by
f(x) = \(\left\{\begin{array}{l}
x+5, \text { if } x \leq 1 \\
x-5, \text { if } x>1
\end{array}\right.\) a continuous funciotn?
(d) Show that
f(x) = \(\left\{\begin{array}{l}
5 x-4, \text { when } 0<x \leq 1 \\
4 x^3-3 x, \text { when } 1<x<2
\end{array}\right.\)
is continuous at x = 1.
Solution:
(a) Given f(x) = \(\left\{\begin{array}{l}
2 x+3, \text { if } x \leq 2 \\
2 x-3, \text { if } x>2
\end{array}\right.\)
Here D_f = R So we check the continous of f at all points of R. Let c ∈ D_f be any arbitrary point.
Three cases arises.
Case-I
when c < 2 then f(c) = 2c + 3
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 2x + 3 = 2c + 3
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = f(c)
Thus f is continous for all c < 2

Case-II
when c > 2 then f(c) = 2c – 3
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 2x – 3 = 2c – 3
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = f(c)
Hence f is continous for all c > 2

Case-III
when c = 2
\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) 2x – 3 = 4 – 3 = 1
and \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) 2x – 3 = 4 + 3 = 7
\(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) ≠ \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\)
Thus f is not continous i.e., discontinous at x = 2. So on contining all these cases, f is continuous at every point of its domain except x = 2.

(b) at x = 0,

Thus \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) does not exists Hence f is discontinuous at x = 0

(c) Given f(x) = \(\left\{\begin{array}{l}
x+5, \text { if } x \leq 1 \\
x-5, \text { if } x>1
\end{array}\right.\)
Hence D_f = R so we examine the continuity of f at all points of R.
Let c ∈ D_f be any arbitrary point so three cases arises.
Case-I
when c < 1 then f(c) = c + 5
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x + 5 = c + 5 = f(c)
∴ f is continuous at all c < 1

Case-II
when c > 2.
Then f(c) = c – 5
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x – 5 = c – 5 = f(c)
∴ f is continuous at all c < 1

Case-III
when c = 1;

Thus f is discontinous at x = 1
Hence on combining all three cases, f is not continuous at every point of its domain.
∴ f is not a continuous function.

Thus f is continuous at x = 1

Question 7.
Examine the continuity for the following functions:
(a) f(x) = \(\begin{cases}x+1, & \text { if } x \geq 1 \\ x^2+1, & \text { if } x<1\end{cases}\)
(b) f(x) = \(\begin{cases}x^{10}-1, & \text { if } x \leq 1 \\ x^2, & \text { if } x>1\end{cases}\)
Solution:
(a) Given f(x) = \(\begin{cases}x+1, & \text { if } x \geq 1 \\ x^2+1, & \text { if } x<1\end{cases}\)
Here D_f = R. so we examine the continuity of f at all x ∈ R. Let c ∈ R be any real number.
Then three cases arises.
Case-I when c < 1 then f(c) = c² + 1
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x² + 1 = c² + 1 = f(c)
Thus f is continous at all c < 1

Case-II when c > 1 Then f(c) = c + 1
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x + 1 = c + 1 = f(c)
∴ f is continuous for all c > 1

Case-III when c = 1;

Thus f is continuous at c = 1
So on combining all three cases, f is continuous at every point of its domain
Hence f be a continuous function.

Question 8.
Discuss the continuity of the function f(x) at x = 1/2 when f(x) is defined as follows :
f(x) = \(\left\{\begin{array}{c}
\frac{1}{2}+x, 0 \leq x<\frac{1}{2} \\
1, x=\frac{1}{2} \\
\frac{3}{2}+x, \frac{1}{2} \end{array}\right.\)
Solution:

Question 9.
Examine the continuity of the function f(x) = \(\left\{\begin{array}{c}
-2, \text { if } x \leq-1 \\
2 x, \text { if }-11
\end{array}\right.\)
Solution:
f(x) = \(\left\{\begin{array}{c}
-2, \text { if } x \leq-1 \\
2 x, \text { if }-11
\end{array}\right.\)
Here D_f = R
So we examine the continuity of f at all x ∈ R. Let c ∈ D_f be any
Case-I
when c < – 1 then f(c) = – 2
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) – 2 = – 2 = f(c)
∴ f is continuous at all c < – 1

Case-II
when – 1 < c > 1
Then f(c) = 2c
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 2x = 2c = f(c)
∴ f(x) is continuous at x = c, where – 1 < c > 1

Case-III
when c > 1
Then f(c) = 2
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f (x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 2 = 2 = f(c)
Thus f is continuous for all c > 1

Case-IV
when c = – 1

Thus f is continuous at x = 1
Thus on combining all five cases, f is continuous at every point of its domain.. Hence / be a continuous function.

Question 10.
A function f (x) is defined as follows: f(x) = xcos\(\frac { 1 }{ x }\), when x ≠ 0, f(0) = 0.
Examine the continuity at x = 0.
Solution:
Given f(x) = \(\begin{cases}x \cos \frac{1}{x} ; & x \neq 0 \\ 0 ; \quad x=0\end{cases}\)
Let h(x) = x ; g(x) = cos \(\frac { 1 }{ x }\)
\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) h(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x = 0
Since cos\(\frac { 1 }{ x }\) is bounded in the deleted x ngd of 0 i.e., \(\left|\cos \frac{1}{x}\right|\) ≤ 1
i.e., cos\(\frac { 1 }{ x }\) is oscillating between – 1 and 1
∴ g(x) be a bounded function

∴ f is continuous at x = 0

Question 11.
The function f(x) is defined as follows:
f(x) = \(\begin{cases}(x-a) \cos \frac{1}{x-a}, & \text { when } x \neq a \\ 0, & \text { when } x=a\end{cases}\)
Examine the continuity, when x = a.
Solution:
Given

Since cos \(\frac{1}{x-a}\) is bounded function in the deleted neighbourhood of point x = a
Also, \(\left|\cos \frac{1}{x-a}\right|\) ≤ 1
i.e, cos\(\frac{1}{x-a}\) is oscilating between -1 & 1.
Thus g(x) be a bounded function.

Thus f is continuous at x = a

Question 12.
Examine the continuity of the following functions:
(a) f(x) = \(\left\{\begin{array}{r}
\frac{|x|}{x}, \text { if } x \neq 0 \\
0, \text { if } x=0
\end{array}\right.\)
(b) f(x) = \(\left\{\begin{array}{l}
\frac{|x|}{x}, \text { if } x<0 \\ -1, \text { if } x \geq 0 \end{array}\right.\)
Solution:
(a)

So we examine the continuity of f at all x ∈ R.
Let c ∈ R be any real number
Case-I when c > 0 Then f(c) = 1

Thus f is discontinuous at x = 1
Therefore on combining all three cases, f is continuous at every point of its domain except at x = 0

(b) Given

Thus f is continuous at x = 0

Question 13.
Examine the continuity at x = 0
(a) f(x) = \(\left\{\begin{array}{c}
\frac{\tan 2 x}{3 x}, \text { when } x \neq 0 \\
\frac{2}{3}, \text { when } x=0
\end{array}\right.\)
(b) f(x) = 1 + \(\frac{|x|}{x}\) for x ≠ 0 and f(0) = 1.
Solution:

Question 14.
Find the value of value of a, if the function f (x) defined by
f(x) = \(\left\{\begin{array}{r}
2 x-1, x<2 \\ a, x=2 \\ x+1, x>2
\end{array}\right.\)
is continuous at x = 2.
Solution:

Question 15.
Find the relationship between a and 6 so that the function defined by f(x) = \(\left\{\begin{array}{l}
a x+1, \text { if } x \leq 3 \\
b x+3, \text { if } x>3
\end{array}\right.\) is continuous at x = 3.
Solution:
Given

which is the required relation between a & b.

Question 16.
For what value of X is the function f(x) = \(\begin{cases}\lambda\left(x^2-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0\end{cases}\) continous at x = 0?
Solution:
at x = 0

∴ L.H.Limit ≠ R.H.Limit
Thus f is not continuous at x = 0 i.e.
at any real value of λ.
at x = 1 :
\(\underset{x \rightarrow 1}{\mathrm{Lt}}\)f(x)= \(\underset{x \rightarrow 1}{\mathrm{Lt}}\)4x+ 1 = 4 + 1 =5 = f(1)
∴ f is continuous at x = 1

Question 17.
Find the value of k, so that the function f defined by
f(x) = \(\begin{cases}k x+1, & \text { if } x \leq \pi \\ \cos x, & \text { if } x>\pi\end{cases}\)
is continuous at x = π.
Solution:

Question 18.
Find the value of k, for which
f(x) = \(\begin{cases}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} & \text { if }-1 \leq x<0 \\ \frac{2 x+1}{x-1}, & \text { if } 0 \leq x<1\end{cases}\)
is continuous at x = 0.
Solution:
at x = 0

Question 19.
If the following fimctoin f(x) is continuous at x = 0, then find the value of k.
f(x) = \(\left\{\begin{array}{cc}
\frac{1-\cos 2 x}{2 x^2}, & x \neq 0 \\
k, & x=0
\end{array}\right.\)
Solution:

Question 20.
Prove that the funciton
f(x) = \(\left\{\begin{array}{cc}
\frac{x}{|x|+2 x^2}, & x \neq 0 \\
k, & x=0
\end{array}\right.\)
remains discontinuous at x = 0, regardless of the choice of k.
Solution:

Thus f is remains discontinuous at x = 0 regardless of the choice of k i.e. whatever the real value of k may be. On removable discontinuity;

Question 21.
The function f(x) = \(\frac{2 x^2-8}{x-2}\) is undefined at x = 2. What value should be assigned to f (2) so that f (x) is continuous at x = 2 ?
Solution:

Since f(x) is continuous at x = 2
if \(\underset{x \rightarrow 2}{\mathrm{Lt}}\)f(x) = f(2) if 8 = f(2)
Thus required value of f(2) which is assigned be 8.

Question 22.
The function f(x) = \(\frac{x^2-1}{x^3-1}\) at the point x = 1 ; what should be the value of f (1) such that f (x) may be continuous at x = 1?
Solution:

Now f(x) may be continuous at x = 1
if \(\underset{x \rightarrow 1}{\mathrm{Lt}}\)f(x) = f(1) if \(\frac { 2 }{ 3 }\) = f(1)

Question 23.
Is the function f(x) = \(\frac{3 x+2 \sin x}{x}\) continuous at x = 0 ? If not, how may the function be defined at x = 0 to make it continuous at that point ?
Solution:
Given f(x) = \(\frac{3 x+2 \sin x}{x}\)
Since f(o) be indeterminate form i.e. (\(\frac { 0 }{ 0 }\)) form.
Thus f(x) is not defined at x = 0
∴ f(x) is not continuous at x = 0

∴ f(x) is continuous at x = 0.

On differentiability

Question 24.
Show that the function f(x)=|x – 3|, x ∈ R, is continuous at x = 3 but not differentiable x = 3.
Solution:
Given

∴ f is continuous at x = 3
∴ f is continuous at x = 30
Differentiability at x = 3

Thus, f is not differentiable x = 3
Hence f is continuous but not differentiable at x = 3

Question 25.
Show that the function
f(x) = |x-1| + |x + 1| for all x ∈ R, is not differentiable at the points x = – 1 and x = 1.
Solution:
f(x) = |x-1| + |x + 1|

Question 26.
Show that the function
f(x) = \(\begin{cases}x-1, & \text { if } x<2 \\ 2 x-3, & \text { if } x \geq 2\end{cases}\)
is not differentiable at x = 2.
Solution:

Question 27.
Show that f(x) = | x – 20| is continuous at x = 20 but f'(x) does not exist at x = 20.
Solution:

Examples

Question 1.
A real valued function f is continuous at a point x = a if it is defined at x = a and iff \(\lim _{x \rightarrow a^{-}}\) f(x) = \(\lim _{x \rightarrow a^{+}}\) f(x)
Solution:
f(a) [∵ value of limit = value of function at the same point]

Question 2.
A real function f (x) is said to be differentiable at x = a, if …………….
Solution:
Rf ‘ (a) = Lf ‘ (a)

Question 3.
If u and v are two function of x, then derivative of \(\frac { u }{ v }\) i.e., (\(\frac { u }{ v }\))’ = ……………..
Solution:
\(\left(\frac{u}{v}\right)^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^2}\) ; v ≠ 0

Question 4.
If the graph of a function/can be drawn around a point without lifting the pen from the paper, then the function is ……………..
Solution:
Continuous

Question 5.
If the graph of a function around a point cannot be drawn without lifting the pen from the paper, then function is ……………..
Solution:
Discontinuous

Question 6.
Examine the continuity of the function.
f(x) = \(\left.\begin{array}{c}
x^2 \text { when } x \neq 1 \\
2 \text { when } x=1
\end{array}\right\}\) at x = 1.
The function is ………………. at x = 1.
Solution:
\(\underset{x \rightarrow 1}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1}{\mathrm{Lt}}\)x² = 1² = 1
and f(1) = 2
∴ \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) f(x) ≠ f(1)
∴ The function is discontinuous at x = 1.

Question 7.
The graph of a function has a …………….. as shown in fig. and so it is a …………….. function.

Solution:
Jump, discontinuous

Question 8.
The graph of a function has a …………….. as shown in fig. and so it is a …………….. function.
Solution:
hole, discontinuous

Question 9.
The function f(x) = x³ – 7x² + 5 being a …………….. function is ……………..
Solution:
polynomial, continuous Tick the correct answer.

Question 10.
The composition gof of two continuous functions is continuous/discontinuous.
Solution:
continuous

Question 11.
f(x) = \(\left\{\begin{array}{c}
3 x,-8, \text { if } x \leq 5 \\
2 k, \text { if } x>5
\end{array}\right.\) is continuous, find k
(a) \(\frac { 3 }{ 7 }\)
(b) \(\frac { 7 }{ 2 }\)
(c) \(\frac { 2 }{ 7 }\)
(d) \(\frac { 4 }{ 7 }\)
Solution:

Question 12.
If f(x) = 2x and g (x) = \(\frac { x² }{ 2 }\) + 1, then which of the following can be a discontinuous function
(a) f(x) + g(x)
(b) f(x) – g(x)
(c) f(x).g(x)
(d) \(\frac{g(x)}{f(x)}\)
Solution:
Given f(x) = 2x; g(x) = \(\frac { x² }{ 2 }\) + 1
For option (A); f(x) + g(x) = 2x + \(\frac { x² }{ 2 }\) – 1,
which is polynomial function and hence continuous everywhere.
For option (B); f(x) – g(x) = 2x – \(\frac { x² }{ 2 }\) – 1,
which is polynomial function in x and hence continuous everywhere.
For option (C); f(x) g(x) = 2x (\(\frac { x² }{ 2 }\) + 1) = x³ + 2x.
which is polynomial in x and hence continuous everywhere.
In option (D); \(\frac{g(x)}{f(x)}=\frac{\frac{x^2}{2}+1}{2 x}=\frac{x^2+2}{4 x}\)
which does not exists at x = 0 and hence discontinuous at x = 0.

Question 13.
f(x) = \(\left\{\begin{array}{c}
3 x,-8, \text { if } x \leq 5 \\
2 k, \text { if } x>5
\end{array}\right.\) is continuous, find k.
(a) \(\frac { 3 }{ 7 }\)
(b) \(\frac { 7 }{ 2 }\)
(c) \(\frac { 2 }{ 7 }\)
(d) \(\frac { 4 }{ 7 }\)
Solution:

Question 14.
The function f (x) = \(\frac{4-x^2}{4 x-x^3}\)
(a) discontinuous at only one point
(b) discontinuous exactly at two points
(c) discontinuous exactly at three points
(d) none of these.
Solution:
Given f(x) = \(\frac{4-x^2}{4 x-x^3}\)
since f(x) is not defined at 4x – x³ = 0
i.e. x(4 – x²) = 0
⇒ x = 0, ±2
Thus f(x) is discontinuous exactly at three points x = 0, ± 2.
Since we know that, f (x) is continuous at x = a, Then f(x) should be defined at x = a
and \(\underset{x \rightarrow a}{\mathrm{Lt}}\)f(x) = f(a)

Question 15.
Let f and g be two real functions at a real number c. Then, which of the following statements is/are true?
I. f+ g is a continuous at x = c.
II. f- g is continuous at x = c.
III. f.g is continuous at x = c.
IV. \(\frac { f }{ g }\) is continuous at x = 0 provided g (x) ≠ 0
(a) I and II
(b) II and III
(c) I, II and III
(d) All
Solution:
by standard results.

Question 16.
A real function f is said to be continuous, if it is continuous at every point in the
(a) domain of f
(b) co-domain of f
(c) range off
(d) none of these
Solution:
(a) domain of f

Question 17.
If f(x) = \(\left\{\begin{array}{rll}
k x^2, & \text { if } & x \leq 2 \\
3, & \text { if } & x>2
\end{array}\right.\)
is continuous at x = 2,
then the value of k is
(a) \(\frac { 4 }{ 3 }\)
(b) – \(\frac { 4 }{ 3 }\)
(c) 3
(d) 4
Solution:

Question 18.
Which of the following functions is/are continuous?
I. constant function
II. polynomial function
III. modulus function
IV. sine function
(a) only I
(b) only II
(c) II, III and IV
(d) All of these
Solution:
(d) All of these

Question 19.
The function f (x) = 2x² + cos x + e^x – 2 is
(a) discontinuous at x = 0
(b) discontinuous at x = π
(c) discontinuous at x = \(\frac { π }{ 2 }\)
(d) continuous at all points
Solution:
Since polynomial, constant, exponential functions are continuous everywhere. Further, cosine function is continuous function.
Also, sum of continuous functions is continuous.
Thus the given function f (x) = 2x² + cos x + e^x – 2
is continuous at all points.

Question 20.
If the function f(x) = \(\left\{\begin{array}{ccc}
\frac{x^2-1}{x-1}, & \text { when } & x \neq 1 \\
k, & \text { when } & x=1
\end{array}\right.\)
is given to be continuous at x = 1, then the value of k is
(a) – 2
(b) 3
(c) \(\frac { 1 }{ 2 }\)
(d) 2
Solution:
Let \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1}{\mathrm{Lt}}\)\(\frac{x^2-1}{x-1}\)
= \(\underset{x \rightarrow 1}{\mathrm{Lt}}\)\(\frac{(x+1)(x-1)}{(x-1)}\) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\)x +1 = 1 + 1 = 2
and f(1) = k
Since f(x) is continuous at x = 1
∴ \(\underset{x \rightarrow 1}{\mathrm{Lt}}\)f(x) = f(1) ⇒ 2 = k

Question 21.
The number of points of discontinuity of/ defined by | x | – | x + | is
(a) 1
(b) 2
(c) 0
(d) None of these
Solution:
f(x) = | x | – | x + 1 |

Thus, f(x) is continuous at x = 0.
Further polynomial function and modulus function are all continuous and difference of two continuous function is continuous. Thus f (x) is continuous everywhere.
∴ The number of points of discontinuity of f be 0.

Question 22.
If a function f is differentiable at c, then it is
(a) discontinuous at c
(b) continuous at c
(c) not defined at c
(d) None of these
Solution:
Given f(x) is differentiable at x = c

Question 23.
Which of the following is not a correct statement?
A function is not differentiable at x = a, if
(a) either or both Rf’ (a) and Lf’ (a) do not exist
(b) both Rf’ (a) and Lf’ exist (a) exist but are not equal
(c) Rf’ (a) and Lf’ (a) both exist and are equal
(d) either or both Rf’ (a) and Lf’ (a) are not finite
Solution:
(c) Rf’ (a) and Lf’ (a) both exist and are equal
[if Lf’ (a) = Rf’ (a) Then f (x) is diff. at x = a]

Question 24.
Let f(x) = \(\left\{\begin{array}{cll}
c x^2+2 x, & \text { if } & x<2 \\
2 x+4, & \text { if } & x \geq 1
\end{array}\right.\). If the function is continuous on (- ∞, ∞), then the value of a is equal to
(a) 4
(b) 2
(c) 3
(d) 1
Solution:

Question 25.
The function f(x) = \(\left\{\begin{array}{cl}
x-1, & x<2 \\
2 x-3, & x \geq 2
\end{array}\right.\) is
continuous function
(a) at x = 2 only
(b) for all integral values of x only
(c) for all real values of x
(d) for all real values of x such that x ≠ 2
Solution:
When x < 2 ; f (x) = x – 1 which is a polynomial function and hence continuous everywhere.
When x > 2 ; f(x) = 2x – 3, which is a polynomial function and hence continuous everywhere, at x = 2

= 4 – 3 = 1 and f (2) = 1
∴ f(x) is continuous at x = 2
Thus f (x) is continuous for all real values of x.

Question 26.
Let f (x) be a function differentiable at x = c, then \(\lim _{x \rightarrow c}\) f(x) equals
(a) f(c)
(b) f ”(x)
(c) \(\frac { 1 }{ f(c) }\)
(d) None of these
Solution:
Given f(x) is differentiable at x = c

Question 27.
If f(x) = \(\left\{\begin{array}{cl}
\frac{1-\cos p x}{x \sin x}, & x \neq 0 \\
\frac{1}{2}, & x=0
\end{array}\right.\)
is continuous at x = 0, then p is equal to
(a) 2
(b) – 2
(c) 1, – 1
(d) none of these
Solution:

Question 28.
The function f (x) = | sin x |
(a) f is differentiable everywhere
(b) f is continuous everywhere but not differentiable at x = nπ, n ∈ Z
(c) f is continuous everywhere but not differentiable at x = (2n + 1) \(\frac { π }{ 2 }\), n ∈ Z
(d) none of these
Solution:
f(x) = | sin x | = \(\left\{\begin{aligned}
-\sin x ; & x \sin x ; & x \geq n \pi
\end{aligned}\right.\)
n = even
at x = nπ (n even)

Thus f(x) is not differentiable at x = nπ (n = odd)
Hence f(x) is not differentiable at x = nπ, n ∈ Z.

Question 29.
If f(x) = \(\left\{\begin{array}{cc}
m x+1, & x \geq \frac{\pi}{2} \\
\sin x+n, & x<\frac{\pi}{2}
\end{array}\right.\) is continuous at x = \(\frac { π }{ 2 }\), then
(a) m = 1, n = 0
(b) m = \(\frac { nπ }{ 2 }\) + 1
(c) n = \(\frac { mπ }{ 2 }\)
(d) m = n = \(\frac { π }{ 2 }\)
Solution:

Question 30.
Determine the value of constant ‘k’ so that the function f (x) = \(\left\{\begin{array}{r}
\frac{k x}{|x|}, x<0 \\
3, x \geq 0
\end{array}\right.\) is continuous at x = 0.
Solution:
At x = 0
L.H.L = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) \(\frac{k x}{-x}\) = – k
R.H.L = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) 3 = 3
and f(0) = 3
Since f(x) is continuous at x = 0.
∴ L.H.L = R.H.L = f(0)
Thus, – k = 3 ⇒ k = – 3

Question 31.
If the function f defined as f(x) = \(\left\{\begin{array}{cc}
\frac{x^2-9}{x-3}, & x \neq 3 \\
k & x=3
\end{array}\right.\)
is continuous at x = 3, find the value of A.
Solution:

Question 32.
If f(x) = \(\left\{\begin{array}{cc}
x^2, & \text { when } x \neq 1 \\
0 & \text { when } x=1
\end{array}\right.\) find whether it is continuous or discontinuous at x = 1.
Solution:
\(\underset{x \rightarrow 1}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) x² = 1 and f (1) = 0
∴ \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) f(x) ≠ f(1)
Thus, f is not continuous at x = 0.

Question 33.
If the function
f(x) = \(\left\{\begin{array}{cc}
\frac{\sin \frac{3 x}{2}}{x}, & x \neq 0 \\
k, & 0
\end{array}\right.\) is continuous
at x = 0, then write the value of A.
Solution:

Question 34.
Examine the continuity of the function f(x) = x³ + 2x² – 1.
Solution:
Given f (x) = x³ + 2x² – 1
Clearly f (x) being a polynomial function and hence continuous at every real point x ∈ R.

Question 35.
Examine the continuity of
f(x) = \(\left\{\begin{array}{rll}
3 x+5, & \text { if } & x \geq 2 \\
x^2, & \text { if } & x<2 \end{array}\right.\)
Solution:
Case-I: When x > 2 ; f(x) = 3x + 5
We know that, every polynomial function is continuous everywhere.
∴ f(x) is continuous at each x > 2.

Case-II: When x < 2 ; f{x) = x², which is polynomial in x and hence continuous everywhere.
Thus, f(x) is continuous at each x < 2.

Case-III : at x = 2
\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) 3x + 5 = 6 + 5 = 11
\(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) x² = 4
∴ \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x)
Thus f (x) is discontinuous at x = 2
Hence f (x) is continuous everywhere except at x = 2.

Question 36.
Examine the continuity of
f(x) = \(\left\{\begin{array}{cl}
\frac{|x-4|}{2(x-4)}, & \text { if } x \neq 4 \\
0, & \text { if } x=4
\end{array} \text { at } x=4\right.\)
Solution:

Question 37.
Examine the continuity
f(x) = |x| + |x – 1|, at x = 1
Solution:
Given f(x) = | x | + | x – 1 |

Thus, f(x) is continuous at x = 1.

Question 38.
If f(x) = 2 | x | + 3 | sin x | + 6, then the right hand derivative off (x) at x = 0 is ……………
Solution:

Question 39.
Let f(x) = x | x |, for all x ∈ R, check its differentiability at x = 0.
Solution:

The availability of step-by-step S Chand Class 12 Maths Solutions  Chapter 25 Application of Integrals Ex 25(b) can make challenging problems more manageable.

S Chand Class 12 ICSE Maths Solutions Chapter 25 Application of Integrals Ex 25(b)

Question 1.
Find the area of the region enclosed by the following curves or lines.
(a) y = 2x, y = x ²
(b) y² = 8x, x² = 8y
(c) y² = x,x² = y
Answer:
(a) The eqns. of given curves are:
y = 2x
and y = x²
eqn. (2) represents an upward parabola with vertex (0,0)
and the line (2) and intersects parabola when 2x = x² =
=> x(x-2) = 0
=> x = 0, 2
When x = 0 ∴ from (1); y = 0
When x = 2 ∴ from (1); y = 4
i. e. points of intersection are (0, 0) and (2, 4).
∴ Required area of shaded region
= \(\int_0^2\) y of line (1) – y of parabola (2) d x
= \(\int_0^2\) 2 x d x – \(\int_0^2\) x² dx
= x²]⁰₂ – \(\frac{x^3}{3}\)₀ ²
= 4 – \(\frac{8}{3}\)
= \(\frac{4}{3}\) sq.units

(b) Given eqns. of parabolas are
x² = 8y …..(1)
and y² = 8x …(2)
both parabolas intersects when \((\frac{x^2}{8})^2\) = 8x
=> x⁴ – 512x = 0
=> x(x³ – 512) = 0
=> x = 0, 8
∴ from (1); y = 0, 8
Thus the points of intersection of both curves are (0, 0) and (8, 8). Divide the region into vertical strips with lower end on x2 = 8y and upper end on y2 = 8* and this rectangle move from x = 0 to x = 8

(c) The given eqns. of curves are;
y² = x
and x² = y
eqn. (1) represents a right handed parabola with vertex (0, 0). eqn. (2) represents an upward parabola with vertex (0, 0). Both curves intersects when x⁴ = x
=> x(x³ – 1) = 0
=> x = 0, 1
when x = 0 ∴ from (2) ; y = 0
when x = 1 ∴ from (2) ; y = 1
∴ both curves intersects at (0, 0) and (1,1) Divide the shaded region into vertical strips. Each vertical strip has lower end on curve (1) and upper end on curve (1).
∴ required area = \(\int_0^1\)(\(\sqrt{x}\) – x²) dx
= \(\frac{2}{3} x^{3 / 2}\) – \(\frac{x^3}{3}\)]⁰₁
= \(\frac{2}{3}\) – \(\frac{1}{3}\)
= \(\frac{1}{3}\) sq. units

Question 2.
Find the area of the figure bounded by the graphs of the functions y = x², y = 2x – x².
eqn. (1) represents an upward parabola.
Answer:
Given eqns of curves are y = x² ….. (1)
and y = 2x – x² ……. (2)
eqn. (1) represents an upward parabola.
With vertex at (0, 0).
eqn. (2) can be written as x² – 2x = -y
x² – 2x + l – l = -y
=> (x- l)² = – (y – 1)
Clearly represents a downward parabola with vertex at (1, 1).
Both parabolas (1) and (2) intersects when x² = 2x – x².
2x² – 2x = 0
=> 2x (x – 1) = 0
=> x = 0, 1
When x = 0 from(l) ; y = 0
When x = 1 from (1) ; y = 1
i.e. points of intersection are (0, 0) and (1,1). Divide the shaded region into vertical strips. Each vertical strip is having lower end on curve (1) and upper end on curve (2).
∴ required area = \(\int_0^1\) [-x² + (2x – x²)] dx
= \(\int_0^1\) (2 x – 2 x²) d x
= x² – \(\frac{2}{3}\) x³]₀ ¹
= 1 – \(\frac{2}{3}\)
= \(\frac{1}{3}\) sq. units

Question 3.
(i) Find the area bounded by the curve x² = 4y and the straight line x = 4y – 2.
(ii) Find the area cut off from the para- bola 4y = 3 × 2 by the line 2y = 3x + 12.
(iii) Find the area of the region included between the parabola y² = x and the x + y = 2.
Answer:
(i) Given curve x² = 4y be an upward parabola with vertex (0,0). The given line x = 4y – 2 meets x-axis at (- 2, 0) and y-axis at (0, \(\frac{1}{2})\)
Both curves and line intersects when (4y – 2)² = 4y
=> 16y² – 16y + 4 = 4y
=> 16y² – 20y + 4 = 0
=> 4y² – 5y + 1 = 0
=> (y-1)(4y-1) = 0
=> y = 1, \(\frac{1}{4}\)
When y = 1 ; x = 2 and
When y = \(\frac{1}{4}\)
∴ x = -1
Thus points of intersection are (2, 1) and
(-1, \(\frac{1}{4}\))

∴ required area = area of shaded region
= \(\int_{-1}^2\) \(\frac{x+2}{4}\) – \(\frac{x^2}{4}\) dx
= \(\frac{1}{4}\) \(\frac{x^2}{2}\) + 2x – \(\frac{x^3}{3}\)₁ ²
= \(\frac{1}{4}\) [2 + 4 – \(\frac{8}{3}\) – \(\frac{1}{2}\) + 2 – \(\frac{1}{3}\)
= \(\frac{9}{8}\) sq. units

(ii) The given curve be 4y = 3x²
=> x² = \( \frac{4}{3}\)y …….(1)
and eqn. of given line be 2y = 3x + 12 …(2)
Curve (1) represents on upward parabola with vertex at (0,0). Line (2) meets x-axis at (- 4,0) and (0, 6).
Thus line (2) meets parabola (1)
When x² = \(\frac{4}{3}\) \(\frac{3 x+12}{2}\)
=> x² = 2 (x + 4)
=> x² – 2x – 8 = 0
=> (x – 4) (x + 2) = 0
=> x= 2, -2
When x = 4 ∴ from (2) ; y = 12
When x = -2 from (2) ; y = 3
Divide the region into vertical strips with ever vertical strip is having lower end on curve (1) and upper end on curve (2).

∴ required shaded area = \(\int_{-2}^4\) \(\frac{3 x+12}{2}\) – \(\frac{3}{4}\) x² d x \(\frac{3}{2}\) \(\frac{(x+4)^2}{2}\) – \(\frac{1}{4}\) x³]_{3 4}
= \(\frac{3}{4}\) × 64 – \(\frac{1}{4}\) × 4³ – \(\frac{3}{4}\) × 2² + \(\frac{1}{4}\)(-2)³
= 48 – 16 – 3 – 2 = 27 sq. units

(iii) We want to find the area of the region included between given two curves
y² = x …(1) and
x + y = 2 ……. (2)
Now eqn. (1) is a parabola with axis x-axis and vertex (0, 0). eqn. (2) represents a line having intercepts on coordinate axes are (2, 0) and (0, 2). eqn. (1) and (2) intersects when y² = 2 – y
=> y² + y – 2= 0
=> y = 1,-2
When y= 1 => x = 2 – 1 = 1
When y = – 2 => x = 4
Thus the points of intersection are (1, 1) and (4, – 2).
We divide the shaded region into horizontal strips. Each horizontal strip is having left end on curve y2 = x and right end on line x + y = 2.
∴ the approximating rectangle was length |X¹ – x²| land width dy

∴ required area = \(\int_{-2}^1\) |x₁ – x₂| dy
= \(\int_{-2}^1\) [(2-y) – y²] d y
= \(\frac{(2-y)^2}{-2}\)]²₁ – \(\frac{y^3}{3}\)² ₁
= \(-\frac{1}{2}\)[1 – 16] – \(\frac{1}{3}\)[1-(-8)]
= + \(\frac{15}{2}\) – 3
= \(\frac{9}{2}\) sq. units

Question 4.
Find the area of the region bounded by the curves y = x² + 2, y = x, x = 0 and .v = 3. Also, sketch the region bounded by these curves.
Answer:
Given curves are y = x² + 2 …(1)
y = x …(2)
and x = 0, x = 1
eqn. (1) represents a parabola with vertex (0, 2).
eqn. (1) and (2) intersects when x = x² + 2 => x² – x + 2 = 0
∴ x does not gives real values.
Thus the line y = x does not meet the parabola y = x² + 2.
Clearly x = 1 meets y = x² + 2 at (1,3).

Divide the shaded region into vertical strips. Each vertical strips has lower end on y = x and upper end on y = x² + 2.
length of approximating rectangle = |y² – y¹| and width dx.
Clearly it moves from x = 0 to x = 1
∴ required area = \(\int_0^1\) [x² + 2 – x] dx
= \(\frac{x^3}{3}\) + 2x – \(\frac{x^2}{2}\)⁰ ₁
= \(\frac{1}{3}\) + 2 – \(\frac{1}{2}\)
= \(\frac{2+12-3}{6}\)
= \(\frac{11}{6}\)sq.units

Question 5.
Find the point of intersection of the liney = 4x with the curve y = x³ If A is that point of intersection which lies in the first quadrant and O is the origin, calculate the area between line OA and the curve.
Answer:
The eqn. of given line be y = 4x …(1)
and eqn. of given curve be y = x³ …….(2)
both curves (1) and (2) intersects when 4x = x³
When x = 0 ∴ from (1); y – 0
When x = ± 2 ∴ from (1); y = ± 8
Since the point of intersection of given line and curve lies in the first quadrant.
∴ Coordinates of A are (2, 8).
Divide the shaded region into vertical strips. Each vertical strip is having lower end on curve (2) and upper end on line (1).

∴ required area of shaded region
= \(\int_0^2\) [4 x – x³] d x
= 2x³ – \(\frac{x^4}{4}\)₀ ²
= (8 – 4) = 4 sq. units

Question 6.
Using integration, find the area of the mangle, whose vertices are
(i) (-1,0), (1,3) and (3,2)
(ii) (2, 5), (4, 7) and (6,2)
Answer:
(i) LetA (-1,0),B (l,3)andC (3,2) are the vertices of ∆ABC
eqn. of line AB be given by, y – 0 = \(\frac{3-0}{1+1}\)(x + 1) => y = \(\frac{3}{2}\)(x + 1)
eqn. of line AB be given by, y – 0 = \(\frac{2-0}{4}\) => y = \(\frac{1}{2}\) (x – 1)
eqn. of line AB be given by, y – 3 = \(\frac{2-3}{3-1}\)(x – 1)
=> y – 3 = \(– \frac{1}{2}\) (x – 1)
=> = \(-\frac{x}{2}\) + \(\frac{7}{2}\)
∴ reqd. area = area of region ABDA + area of region BDCB
= \(\int_{-1}^1\) \(\frac{3}{2}\)(x + 1) – \(\frac{1}{2}\) (x + 1) d x + \(\int_1^3\)\(\frac{-x}{2}\) + \(\frac{7}{2}\) – \(\frac{1}{2}\) x – \(\frac{1}{2}\) dx
= \(\frac{(x+1)^2}{2}\)^-1+1 + \(\int_1^3\)-(x – 3) dx
= \(\frac{1}{2}\)[4 – 0] – \(\frac{1}{2}\)(x – 3)² ]^{1 3}
= 2 – \(\frac{1}{2}\)(0 – 4) = 4 sq. units.

(ii) Let the given vertices of ∆ABC are A (2, 5), B (4, 7) and C (6, 2)
eqn. of line AB be given by
y – 5 = \(\frac{2-7}{6-4}\)(x – 2)
=> y – 5 = x – 2
=> y = x + 3
eqn. of line BC be given by
y – 7 = \(\frac{2-7}{6-4}\)(x- 4)
=> y – 7 = –\(\frac{5}{2}\) (x- 4)
=> 2y – 14 = -5x + 20
=> 2y + 5x = 34

eqn. of line AC be given by
y – 5 = \(\frac{2-5}{6-2}\) (x – 2)
=> y – 5 = \(\frac{-3}{4}\) (x – 2)
=> 4y – 20 = -3x + 6
=> 4y + 3x = 26
∴ area of shaded ∆ABC = area of region AMNB + area of region BNLC area of region AMLC

Question 7.
Using integration, find the area of the region bounded by the triangle whose sides are y = 2x + l,y = 3x + l and x = 4.
Answer:
To find the area of triangular region bounded by curves given as under
y = 2x + 1 …..(1)
y = 3x + 1 …..(2)
and x = 4 …..(3)
On solving (1) and (2); we have 2x + 1 = 3x + 1
=> x = 0
∴ y = 1
Thus line (1) and (2) intersects at (0, 1). eqn. (1) and (3) intersects at (4, 13). and eqn. (1) and (3) intersect at (4, 9). required area = area of ∆ABC.
Divide the region into vertical strips. Each vertical strip has lower end on liney = 2x + 1 and upper end on line y = 3x + 1.
So length of approximating rectangle = |y₁ – y₂| and width = dx

and clearly it moves from 0 to 4.
∴ required area = \(\int_0^4\) |ysub>1 – ysub>1| dx
= \(\int_0^4\)[(3x + 1) – (2x + 1)][/latex] dx
[∵ y₁ ≥ y₂
= \(\frac{x^2}{2}\)⁰₄ = 8 sq. units

Question 8.
Find the area of the region enclosed between two circles x² + y² = 4 and (x- 2)² + y² = 4.
Answer:
The given circles are x² + y² = 4 …(1)
and (x – 2)² + y² = 4 …(2)
eqn. (1) represents a circle with centre at (0, 0) and radius 2 and eqn. (2) represents a circle with centre (2, 0) and radius 2.
Now eqn. (1) and (2) intersects
when (x – 2)² – x² = 0
i.e x² – 4x + 4 – x² = 0 i.e x = 1
∴ y = ±√3
∴ pts. of intersections are (1,±√3).
∴ Required area = 2 [area OACO + area ABCA]
= 2 \(\int_0^1\) \(\sqrt{4-(x-2)^2}\) d x + \(\int_1^2\) \(\sqrt{4-x^2}\) d x
[Region lies in Ist quadrant]

Question 9.
Sketch the region common to the circle x² + y² = 16 and the parabola X² = 6y. Also find the area of the region using integration.
Answer:
Given region be {(x, y): x² + y² ≥ 16 ; x² ≤ 6y}
given eqn. of circle be JC2 +y2 = 16 which represents a circle with centre (0, 0) and radius 4. eqn. of given parabola be x2 = 6y
which represents an upward parabola with vertex at (0, 0). both curves eqn. (1) and eqn. (2) intersects y² + 6y – 16 = 0
(y – 2) (y + 8) = 0
=> y = 2,-8
when y = 2 ∴ from (2); x = ±2√3
Thus both curves intersects at (±2√3, 2)
and y = -8 does not gives any real values of x required area = 2 x area of OABO

Question 10.
Find the area given by x + y ≤ 6, x² + y² <, 6y and y2 ≤ 8x.
Answer:
The eqns. of given curves are
x + y = 6 …(1)
x2² + y² = 6y …(2)
and y² = 8x

lines (1) meets coordinates axes at A (6, 0) and B (0, 6). eqn. (2) represents a circle with centre C (0, 3) and radius 3. eqn. (3) represents a right handed parabola with vertex at O (0, 0). line (1) and circle (2) intersects when (6 – y)² + y² = 6y
=> 2y² – 18y + 36 = 0
=> y² – 9y + 18 = 0
=> (y – 3) (y – 6) = 0
=> y = 3, 6
when y = 3 ∴ from (1) ; X = 3
when y = 6 ∴ from (1) ; X = 0
i.e. points of intersection are (2, 4) and (18, – 12)
Divide the region into horizontal strips.
∴ required area = area of region OFDO + area of region EFDE

Question 11.
Indicate the region bounded by the curve x² = y, y = x + 2 and X-axis and obtain the area enclosed by them.
Answer:
Given eqns. of curves are, x² = y …..(1)
and y = x + 2 …..(2)
eqn. (1) represents a upward parabola with vertex (0, 0).
The line (2) meets coordinate axes at (- 2, 0) and (0, 2).
eqn. (1) and (2) intersects when
x² – x – 2 = 0
(x – 2) (x + 1) = 0 => x = -l, 2
When x = – 1 ∴ from (2); y = 1
When x = 2 ∴ from (2); y = 4
∴ points of intersection are (- 1, 1) and (2, 4).
Divide the region into vertical strips with upper end on line and lower end on parabola.
The length of approximating rectangle = |y² – y¹| and width = dx
Clearly the rectangle moves from x = – 1 to x = 2.

∴ required area = \(\int_{-1}^2\) |y₂ – y₁| d x
= \(\int_{-1}^2\) (y₂ – y₁) d x
= \(\int_{-1}^2\) [x + 2 – x²] d x
= [\(-\frac{x^2}{2}\) + 2 x – \(\frac{x^3}{3}\)]² ₁
= [2 + 4 – \(\frac{8}{3}\) – \(\frac{1}{2}\) + 2 – \(\frac{1}{3}\)]
= 5 – \(\frac{1}{2}\)
= \(\frac{9}{2}\) sq. units

Question 12.
Using integration, find the area of the triangle formed by positive x-axis and tangent and normal to the circle x² + y² = 4 at (1, \(\sqrt{3}\)).
Answer:
Given eqn. of circle be x² + y² = 4 with centre (0, 0) and radius 2 . Diff. given eqn. w.r.t. x; we have
2 x + 2 y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = \(-\frac{x}{y}\)

∴ slope of tangent to given circle at (1, \(\sqrt{3}\))
= \(\frac{-1}{\sqrt{3}}\)
Thus, eqn. of tangent to given circle at (1, \(\sqrt{3}\)) is given by
y – \(\sqrt{3}\) = \(-\frac{1}{\sqrt{3}}\)(x – 1)
⇒ x + \(\sqrt{3}\) y = 4 …………..(1)
∴ slope of normal to given circle at (1, \(\sqrt{3}\))
= \(\sqrt{3}\)
∴ eqn. of normal to given circle at point (1, \(\sqrt{3}\)) be given by
y – \(\sqrt{3}\) = \(\sqrt{3}\)(x – 1)
⇒ \(\sqrt{3}\) x – y = 0
line (1) meets coordinate axes at (4, 0) and (0, \(\frac{4}{\sqrt{3}}\)).
∴ required area = area of region OACO + area of region ACBA
= \(\int_0^1\) \(\sqrt{3}\) x d x + \(\int_1^4\) \(\frac{4-x}{\sqrt{3}}\) dx
= \(\sqrt{3}\) \(\frac{x^2}{2}\)]¹ ₀ + \(\frac{1}{\sqrt{3}}\) \(\frac{(4-x)^2}{-2}\)]⁴ ₁
= \(\frac{\sqrt{3}}{2}\) – \(\frac{1}{2 \sqrt{3}}\)[0 – 9]
= \(\frac{\sqrt{3}}{2}\) + \(\frac{9}{2 \sqrt{3}}\)
= \(\frac{\sqrt{3}}{2}\) + \(\frac{3 \sqrt{3}}{2}\)
= \(\frac{4 \sqrt{3}}{2}\)
= 2 \(\sqrt{3}\) sq. units

examples:

Question 1.
Calculate the area of the figure bounded by the curvey = log x, the straight line x = 2 and the .Y-axis,
Answer:
Given eqn. of given curve y = log x
The curve meets .v-axis i.e. y = 0
∴ loge x – 0
=> x = e⁰ = 1
Thus required area = \(\int_1^2\) y dx = \(\int_1^2\) log x dx = \(\int_1^2\) log x . 1 dx
= logx – x]² ₁ – \(\int_1^2\) \(\frac{1}{x}\) xdx
= xlogx – x]² ₁
= (2 log 2 – 2) – (1 log 1 – 1) = 2 log 2 – 2 – 0 + 1
= 2 log 2 – 1
= log 2² – log e
= log \(\frac{4}{e}\) sq units

Question 2.
Find the area of the figure bounded by the graphs of the function y = x² and y = 2x – x².
Answer:
Given eqns. of curves are y = x² …(1)
and y = 2x – x² …(2)
eqn. (1) represents an upward parabola.
With vertex at (0, 0). eqn. (2) can be written as x² – 2x = -y
x² – 2x + 1 – 1 = – y
=> (x – 1)² = -(y – 1)
Clearly represents a downward parabola with vertex at (1,1).
Both parabolas (1) and (2) intersects when x² = 2x – x²
2x² – 2x = 0
=> 2x(x – 1) = 0 => x = 0, 1

When x = 0 .’. from(1); y = 0
When x = 1 from (1); y = 1
i.e. points of intersection are (0, 0) and (1, 1).
Divide the shaded region into vertical strips.
Each vertical strip is having lower end on curve (1) and upper end on curve (2).
∴ required area = \(\int_0^1\) [-x² + (2x – x²)] dx
= \(\int_0^1\) (2x – x²)dx
= x² – \(\frac{2}{3}\)x³]¹ ₀
= 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\) sq.units

Question 3.
Show that the area ihehwled between the .Y-axis and the curve a²y = x²(x + a) is \(\frac{a² }{12}\)
Answer:
Clearly the curve a²y = x² (x + a)
meets x-axis i.e. y- 0, where x² (x + a) = 0
=> x = 0,-a
∴ curve meets x-axis at (0, 0) and (- a, 0).
∴ required area = \(\int_{-a}^0\) y d x
= \(\int_{-a}^0\) \(\frac{1}{a^2 }\)(x³ + a x²) d x
= \(\frac{1}{a^2 }\) [ \(\frac{x^4}{4}\) + \(\frac{a x^3}{3}\)]^-a ₀
= \(\frac{1}{a^2}\) [0 + 0 – \(\frac{a^4}{4}\) + \(\frac{a^4}{3}\)
= \(\frac{a^2 }{12}\) sq. units

Question 4.
Sketch and shade the area of the region lying in the first quadrant and bounded by y = 9 x², x = 0, y = 1 and y = 4. Find the area of the shaded region.
Answer:
Given eqn. of curve be y = 9 x² which represents an upward parabola with vertex at origin (0, 0). Divided the shaded region into horizontal strips. Each horizontal strip is having left end on y-axis and right end on given curve.
∴ Required area = \(\int_1^4\) x d y
= \(\int_1^4\) \(\frac{\sqrt{y}}{3}\) d y
= \(\frac{1}{3}\) \(\frac{y^{3 / 2}}{3 / 2}\)]¹ ₄
= \(\frac{2}{9}\)[8 – 1]
= \(\frac{14}{9}\) sq. units

Question 5.
Calculate the area bounded by the curve y = x(2 – x) and the lines x = 0, y = 0, x = 2.
Answer:
The eqn. of given curve be y = x(2 – x)
⇒ y = 2x – x² = -(x² – 2 x + 1 – 1)
⇒ y = -(x – 1)² + 1
⇒ (x – 1)² = -(y – 1)
which represents a downward parabola with vertex (1, 1). eqn. (1) meets x-axis i.e. y = 0
⇒ x = 0, 2 i.e. at points (0, 0) and (2, 0)
divide the shaded region into vertical strips. Each vertical strip is having lower end on x-axis and upper end on given curve 1 .
∴ required area = \(\int_0^2 \) y d x
= \(\int_0^2 \) (2 x – x²) d x
= x² – \(\frac{x^3}{3}\)]²₀
= 4 – \(\frac{8}{3}\)
= \(\frac{4}{3}\) sq. units

Question 6.
Find the area enclosed by the curves y² = x and y² = 4 – 3 x.
Answer:
The equations of given curves are
y² = x
y² = 4 – 3 x
and
eqn. (1) represents a right handed parabola with vertex O(0, 0).
eqn. (2) can be written as; y² = -3(x – \(\frac{4}{3}\))
which represents a left handed parabola with vertex at (\(\frac{4}{3}\), 0).
Both parabolas (1) and (2) intersects when 4 – 3 x = x
⇒ x = 1
∴ from (1); y² = 1
⇒ y = 1
Thus their points of intersections are (1, ± 1).
required area = 2 [area of region OABO]
= 2\(\int_0^1\) \(\sqrt{x}\) d x + \(\int_1^{4 / 3}\) \(\sqrt{4-3 x}\) d x
= 2\(\frac{2 x^{3 / 2}}{3}\)}¹
₀ + \(\frac{2}{3}\) \(\frac{(4 – 3 x)^{3 / 2}}{-3}\)^4/3 ₁
= 2\(\frac{2}{3}\)(1 – 0) – \(\frac{2}{9}\)(0 – 1)
= 2\(\frac{2}{3}\) + \(\frac{2}{9}\)
= \(\frac{16}{9}\) sq. units

Question 7.
Draw a rough sketch of the curve y² + 1 = x, x<2. Find the area enclosed by the curve and the line x = 2.
Answer:
Given equation of curve be y² + 1 = x
⇒ y² = x – 1
Clearly it represents a parabola (right handed) with vertex (1, 0) and does not meeting y-axis at any point.
Further the given curve meets the line x = 2 at y² + 1 = 2
⇒ y = ± 1 i.e. at points (2, + 1) and (2, – 1).
Clearly the given curve is symmetrical about x-axis.
∴ required area = 2 × area of region enclosed by parabola and line x = 2 in first quadrant.
Divide this region R into vertical strips with lower end on x-axis and upper end on y = \(\sqrt{x-1}\)
and corresponding rectangle move from x = 1 and x = 2
[∵ y ≥ 0 ∴ |y| = y]
∴required area
= 2 \(\int_1^2\)|y| d y
= 2 \(\int_1^2\) y d x
= 2 \(\int_1^2\) \(\sqrt{x-1}\) d x
= 2 \(\frac{(x-1)^{3 / 2}}{3 / 2}\)¹ ₂
= \(\frac{4}{3}\)[1 – 0] = \(\frac{4}{3}\)

Question 8.
Draw a rough sketch of the curve x² + y = 9 and find the area enclosed by the curve, the x axis and the lines x + 1 = 0 and x – 2 = 0.
Answer:
Given eqn. of curve be x² + y = 9
⇒ x² = -(y – 9)
which represents a downward parabola with vertex (0, 9).

The given lines x + 1 = 0 i.e. x = -1 and x – 2 = 0 i.e. x = 2
Further given curve meets x-axis at (± 3, 0)
∴ Required area
= \(\int_{-1}^2\) y d x
= \(\int_{-1}^2\) (9 – x²) d x
= 9 x – \(\frac{x^3}{3}\)]²₁
= (18 – \(\frac{8}{3}\)) – (-9 + \(\frac{1}{3}\))
= 18 – \(\frac{8}{3}\) + 9 – \(\frac{1}{3}\)
= 27 – 3 = 24 sq. units

Question 9.
Draw a rough sketch of the curve y = x² – 5 x + 6 and find the area bounded by the curve and the x-axis.
Answer:
Given eqn. of curve be y = x² – 5 x + 6
i.e. y = x² – 5 x + \(\frac{25}{4}\) – \(\frac{25}{4}\) + 6
⇒ y = (x – \(\frac{5}{2}\))² – \(\frac{1}{4}\)
⇒ t(x – \(\frac{5}{2}\))²
= y + \(\frac{1}{4}\)
which represents an upward parabola with vertex (\(\frac{5}{2}\),- \(\frac{1}{4}\)).

The given curve meets x-axis i.e. y = 0
∴ from (1); we have x² – 5 x + 6 = 0
⇒ x = 2,3 i.e. at points (2, 0) and (3, 0)
and given curve meets y-axis at (0, 0)
∴ required area = \(\int_2^3\) |y| d x
= \(\int_2^3\) – (x² – 5 x + 6) d x
= –[\(\frac{x^3}{3}\) – \(\frac{5 x^2}{2}\) + 6 x]² ₃
= [9 – \(\frac{45}{2}\) + 18 – \(\frac{8}{3}\) + 10 – 12]
= -[25 – \(\frac{45}{2}\) – \(\frac{8}{3}\)]
= –\(\frac{150-135-16}{6}\)
= \(\frac{1}{6}\) sq. units

Question 10.
Draw a rough sketch of the curves y = (x – 1)² and y = |x – 1|. Hence, find the area of the region bounded by these curves.
Answer:
Given eqns. of curves are ; y = (x – 1)²
and
y = |x – 1|
eqn. (1) represents an upward parabola with vertex (1,0) and this parabola meets y-axis at (0,1).
Now y = |x – 1| = {x-1 ; x ≥ 1 -(x – 1) ; x < 1}.
Clearly the lines y = x – 1 meets the coordinate axes at (1, 0) and (0, -1).
and The line y = -x + 1 meets coordinate axes at (1, 0) and (0, 1).
Clearly the line y = x – 1 and y = (x – 1)²intersects when x – 1 = (x – 1)²
⇒ (x – 1)(x – 2) = 0
⇒ x = 1, 2
i.e. y = 0, 1 ; i.e. at points (1, 0) and (2, 1)
The line y = -(x – 1) meets the curve y = (x – 1)² when -(x – 1) = (x – 1)²
⇒ x = 1, 0
i.e. y = 0, 1 ie. at points (0, 1) and (1, 0).

Clearly required area = area of shaded region
= 2 \(\int_1^2\) [(x – 1) – (x + 1)²] d x
= 2\(\frac{(x-1)^2}{2}\) – \(\frac{(x-1)^3}{3}\)]₁₂
= 2\(\frac{1}{2}\) – \(\frac{1}{3}\) – 0 + 0
= \(\frac{1}{3}\) sq. units

Question 11.
Find the area of the region bounded by the curve x = 4 y – y² and the y-axis.
Answer:
Given eqn. of curve be
⇒ x = -(y¹ – 4 y + 4 – 4)
⇒ x = -(y – 2)¹ + 4
⇒ (y – 2)¹ = -(x – 4)
which represents a lift handed parabola with vertex (4, 2). Clearly eqn. (1) passes through O(0, 0). eqn. (1) meets y-axis at x = 0
∴ 4 y – y¹ = 0
⇒ y(4 – y) = 0
⇒ y = 0, 4
i.e. at points (0, 0) and (0, 4)

Divide the shaded region into horizontal strips.
Each horizontal strip is having left end on y-axis and right end on given curve.
∴ Required area = \(\int_0^4\) x d y.
= \(\int_0^4\) (4 y – y¹) d y
= \(\frac{4 y^2}{2}\) – \(\frac{y^3}{3}\)]₀ ⁴
= 32 – \(\frac{64}{3}\)
= \(\frac{32}{3}\) sq. units

Question 12.
Find the area bounded by the curve y = 2 x – x² , and the line y = x.
Answer:
Given eqn. of curve be y = 2x – x²
and eqn. of given line be y = x
eqn. (1) can be written as y = -(x² – 2 x + 1 – 1)
⇒ y = -(x – 1)² + 1
⇒ (x – 1)² = -(y – 1)
which represents a downward parabola with vertex at (1, 1). Given curve (1) meets x-axis at y = 0
∴ 2 x – x^2=0
⇒ x = 0, 2
i.e. at points (0, 0) and (2, 0).
Line (2) meets parabola (1) when x = 2 x – x²
⇒ x² = x
⇒ x = 0, 1
∴ from (2); y = 0, 1
i.e. both curves intersects at (0, 0) and (1, 1).
Divide the shaded region into vertical strips. Each strip is having upper end on curve (1) and lower end on line (2).

∴ Required area = \(\int_0^1\) [2x – x² – x] d x .
= \(\int_0^1\)(x – x²) d x
= \(\frac{x^2}{2}\) – \(\frac{x^3}{3}\)_{0 1}
= \(\frac{1}{2}\) – \(\frac{1}{3}\)
= \(\frac{1}{6}\) sq. units

Question 13.
Find the smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2.
Answer:
Given eqn. of circle be
x² + y² = 4
Now eqn. (1) represents a circle with centre (0, 0) and radius 2
and eqn. of given line be x + y = 2 meets coordinates axes at (2, 0) and (0, 2).
∴ required area = area of shaded region ABCA
Divide the region into vertical strips with lower end on line x + y = 2 and upper end on x² + y² = 4. Clearly the rectangle move from x = 0 to x = 2

∴ required area = \(\int_0^2\) \(\sqrt{4-x^2}\) – (2 – x) d x
= \(\frac{x \sqrt{4-x^2}}{2}\) + \(\frac{4}{2}\) sin^-1 \(\frac{x}{2}\)₀ ² – \(\frac{(2-x)^2}{-2}\)² ₀
= [0 + 2 sin^-1(1) – 0 – 0] + \(\frac{1}{2}\)[0 – 4]
= 2 × \(\frac{\pi}{2}\) – 2
= (π- 2) sq. units
∴ Ans. (b)

Question 14.
Find the area of the region bounded by the curves y = 6x – x² and y = x² – 2 x.
Answer:
The given eqns. of curves are
and
y = 6x – x²
y = x² – 2 x
eqn. (1) can be written as y = -(x² – 6 x + 9 – 9)
⇒ y = -(x – 3)² + 9
⇒ (x – 3)² = -(y – 9)
which represents a downward parabola with vertex at (3, 9) and meets x-axis at y = 0
∴ from (1); 6 x – x² = 0
⇒ x(6 – x) = 0 ⇒ x = 0, 6
i.e. curve (1) meets x-axis at (0, 0) and (6, 0) eqn. (2) can be written as;
y = x² – 2 x
⇒ y = x² – 2 x + 1 – 1
⇒ y = (x – 1)² – 1
⇒ (x – 1)² = y + 1
which represent an upward parabola with vertex (1, -1) and meets x-axis at 0 = x² – 2 x i.e. x = 0, 2
i.e. curve (2) meets x-axis at (0, 0) and (2, 0)
both parabolas intersects when

6 x – x² = x² – 2 x
⇒ 2 x² – 8 x = 0 arrow x = 0, 4
When x = 0 ∴ from (1); y = 0
When x = 4 ∴ from (2); y = 24 – 16 = 8
Thus both parabolas intersects at (0, 0) and (4, 8).
Divide the region into vertical strips. Each vertical strip is having upper end on curve (1) and lower end on parabola (2).
∴ Required area = \(\int_0^4\) [(6 x – x²) – (x² – 2 x) d x
= \(\int_0^4\) (8 x – 2 x²) d x
= \(\frac{8 x^2}{2}\) – \(\frac{2 x^3}{3}\)₀⁴
= 4 × 4² – \(\frac{2}{3}\) × 4³
= 64 – \(\frac{128}{3}\)
= \(\frac{64}{3}\) sq. units

The availability of step-by-step ISC S Chand Maths Class 12 Solutions Chapter 6 Matrices Ex 6(j) can make challenging problems more manageable.

S Chand Class 12 ICSE Maths Solutions Chapter 6 Matrices Ex 6(j)

Solve the following systems of equations by matrix method.

Question 1.
2x – 3y = 1
3x – 2y = 4
Solution:
The give system of eqn’s is equivalent to AX = B
where A = \(\left[\begin{array}{ll}
2 & -3 \\
3 & -2
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
& B = \(\left[\begin{array}{l}
1 \\
4
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ll}
2 & -3 \\
3 & -2
\end{array}\right|\) = – 4 + 9 = 5 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
Also given system of equations has unique solution.
The cofactors of R₁ are ; – 2 ; – 3
The cofactors of R₂ are ; 3 ; 2
∴ adj A = \(\left[\begin{array}{cc}
-2 & -3 \\
3 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{ll}
-2 & 3 \\
-3 & 2
\end{array}\right]\)
Thus, A^-1 = \(\frac{1}{5}\left[\begin{array}{ll}
-2 & 3 \\
-3 & 2
\end{array}\right]\)
Since AX = B ⇒ X = A^-1B
⇒ \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\frac{1}{5}\left[\begin{array}{ll}
-2 & 3 \\
-3 & 2
\end{array}\right]\left[\begin{array}{l}
1 \\
4
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}
-2+12 \\
-3+8
\end{array}\right]\)
= \(\frac{1}{5}\left[\begin{array}{c}
10 \\
5
\end{array}\right]=\left[\begin{array}{l}
2 \\
1
\end{array}\right]\)
Thus, x = 2 ; y = 1

Question 2.
2x + 3y = 23
3x + 4y = 32
Solution:
The give system of eqn’s is equivalent to AX = B
where A = \(\left[\begin{array}{ll}
2 & 3 \\
3 & 4
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
& B = \(\left[\begin{array}{l}
23 \\
32
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ll}
2 & 3 \\
3 & 4
\end{array}\right|\) = 8 – 9 = 1 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
Also given system of equations has unique solution.
The cofactors of R₁ are ; 4 ; – 3
The cofactors of R₂ are ; – 3 ; 2
∴ adj A = \(\left[\begin{array}{cc}
4 & -3 \\
-3 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
4 & -3 \\
-3 & 2
\end{array}\right]\)
∴ A^-1 = \(\frac{1}{-1}\left[\begin{array}{cc}
4 & -3 \\
-3 & 2
\end{array}\right]=\left[\begin{array}{cc}
-4 & 3 \\
3 & -2
\end{array}\right]\)
Since AX = B ⇒ X = A^-1B
⇒ X = \(\left[\begin{array}{cc}
-4 & 3 \\
3 & -2
\end{array}\right]\left[\begin{array}{l}
23 \\
32
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
-92+96 \\
69-64
\end{array}\right]=\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
Thus, x = 4 ; y = 5 be the sequared solution.

Question 3.
3x + y + z = 3
2x – y – z = 2
– x – y + z – 1
Solution:
The give system of eqn’s is equivalent to AX = B
where A = \(\left[\begin{array}{ccc}
3 & 1 & 1 \\
2 & -1 & -1 \\
-1 & -1 & 1
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
& B = \(\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ccc}
3 & 1 & 1 \\
2 & -1 & -1 \\
-1 & -1 & 1
\end{array}\right|\)
Expanding along R₁,
= 3(- 1 – 1) – 2(2 – 1) + 1(- 2 – 1)
= – 6 – 1 – 3 = – 10 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A …(1)
Thus, given system of eqn’s has unique solution.
The cofactors of R₁ are ;
\(\left|\begin{array}{cc}
-1 & -1 \\
-1 & 1
\end{array}\right| ;-\left|\begin{array}{cc}
2 & -1 \\
-1 & 1
\end{array}\right| ;\left|\begin{array}{cc}
2 & -1 \\
-1 & -1
\end{array}\right|\)
i.e. – 2 ; – 1 ; – 3
The cofactors of R₂ are ;
–\(\left|\begin{array}{cc}
1 & 1 \\
-1 & 1
\end{array}\right| ;\left|\begin{array}{cc}
3 & 1 \\
-1 & 1
\end{array}\right| ;\left|\begin{array}{cc}
3 & 1 \\
-1 & -1
\end{array}\right|\)
i.e. – 2 ; 4 ; 2
The cofactors of R₃ are ;

i.e. x = 1 & y = – 1 & z = 1
Hence the required solution of given system of eqns. be, x = 1 & y = – 1 & z = 1

Question 4.
x – y – z = 2
2x – y = 0
2y – z = 1
Solution:
The give system of eqn’s is equivalent to AX = B
where A = \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & -1 & 0 \\
0 & 2 & -1
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
& B = \(\left[\begin{array}{l}
2 \\
0 \\
1
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ccc}
1 & -1 & 1 \\
2 & -1 & 0 \\
0 & 2 & -1
\end{array}\right|\)
= 1(1 – 0) + 1(- 2 – 0) + 1(4 – 0)
= 1 – 2 + 4 = 3 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
Thus, given system has unique solution.
The cofactors of R₁are ;
\(\left|\begin{array}{cc}
-1 & 0 \\
2 & -1
\end{array}\right| ;-\left|\begin{array}{cc}
2 & 0 \\
0 & -1
\end{array}\right| ;\left|\begin{array}{cc}
2 & -1 \\
0 & 2
\end{array}\right|\)
i.e. 1 ; 2 ; 4
The cofactors of R₂ are ;
\(-\left|\begin{array}{cc}
-1 & 1 \\
2 & -1
\end{array}\right| ;\left|\begin{array}{cc}
1 & 1 \\
0 & -1
\end{array}\right| ;-\left|\begin{array}{cc}
1 & -1 \\
0 & 2
\end{array}\right|\)
i.e. 1 ; – 1 ; – 2
The cofactors of R₃ are ;

Question 5.
\(\frac{2}{x}+\frac{3}{y}+\frac{10}{z}\) = 4
\(\frac{4}{x}-\frac{6}{y}+\frac{5}{z}\) = 1
\(\frac{6}{x}+\frac{9}{y}-\frac{20}{z}\) = 2
Solution:
Putting \(\frac { 1 }{ x }\) = u; \(\frac { 1 }{ y }\) = v & \(\frac { 1 }{ z }\) = w in
2u + 3v + 10w = 4
4u + 6v + 5w = 1
6u + 9v – 20w = 2
The given system of eqn’s is eqrualant to AX = B
where A = \(\left[\begin{array}{ccc}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right]\); X = \(\left[\begin{array}{l}
u \\
v \\
w
\end{array}\right]\)
& B = \(\left[\begin{array}{l}
4 \\
1 \\
2
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ccc}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right|\)
Taking 2, 3 & 5 common from R₁; R₂ & R₃ respectively
= 2 x 3 x 5\(\left|\begin{array}{ccc}
1 & 1 & 2 \\
2 & -2 & 1 \\
3 & 3 & -4
\end{array}\right|\);
Expanding along R₁,
= 30[1(8 – 3) – 1(- 8 – 3) + 2(6 + 6)]
= 30[5 + 11 + 24] = 30 x 40
= 1200 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
Thus, given system has unique solution.
The cofactors of R₁ are ;
\(\left|\begin{array}{cc}
-6 & 5 \\
9 & -20
\end{array}\right| ;-\left|\begin{array}{cc}
4 & 5 \\
6 & -20
\end{array}\right| ;\left|\begin{array}{cc}
4 & -6 \\
6 & 9
\end{array}\right|\)
i.e. 75 ; 110 ; 72
The cofactors of R₂ are ;
\(-\left|\begin{array}{cc}
3 & 10 \\
9 & -20
\end{array}\right| ;\left|\begin{array}{cc}
2 & 10 \\
6 & -20
\end{array}\right| ;-\left|\begin{array}{ll}
2 & 3 \\
6 & 9
\end{array}\right|\)
i.e. 150 ; – 100 ; 0
The cofactors of R₃ are ;
\(\left|\begin{array}{cc}
3 & 10 \\
-6 & 5
\end{array}\right| ;-\left|\begin{array}{cc}
2 & 10 \\
4 & 5
\end{array}\right| ;\left|\begin{array}{cc}
2 & 3 \\
4 & -6
\end{array}\right|\)
i.e. 75 ; 30 ; – 24

Question 6.
x + y = 5
z + y = 7
z + x = 6
Solution:
The give system of eqn’s is equivalent to AX = B
where A = \(\left[\begin{array}{lll}
1 & 1 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
& B = \(\left[\begin{array}{l}
5 \\
7 \\
6
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{lll}
1 & 1 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1
\end{array}\right|\); Expanding along R₁
= 1(1 – 0) – 1(0 – 1) + 0
= 1 + 1 = 2 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
Also, given system of eqn’s has unique solution.

Thus, x = 2 ; y = 3 & z = 4
Hence the reqd. sol of given system be x = 2 ; y = 3 & z = 4

Question 7.
5x – y = – 7
2x + 3z = 1
3y – z = 5
Solution:
The give system of eqn’s is equivalent to AX = B
where A = \(\left[\begin{array}{ccc}
5 & -1 & 0 \\
2 & 0 & 3 \\
0 & 3 & -1
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
& B = \(\left[\begin{array}{c}
-7 \\
1 \\
5
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ccc}
5 & -1 & 0 \\
2 & 0 & 3 \\
0 & 3 & -1
\end{array}\right|\); Expanding along R₁
= 5(0 – 9) + 1(- 2 – 1)
= – 45 – 2 = – 47 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
Also, given system has unique solution.
The cofactors of R₁ are ;

Question 8.
(i) Find the inverse of the matrix \(\left(\begin{array}{rr}
0.8 & -0.6 \\
0.6 & 0.8
\end{array}\right)\) and use it in solving the equations
0.8x₁ – 0.6x₂ – 10, 0.6x₁ + 0.8x₂ = 20.
(ii) Find the inverse of the matrix \(\left(\begin{array}{ll}
6 & 7 \\
4 & 5
\end{array}\right)\), and use it to solve the simultaneous equations 6x + 7y = 2, 4x + 5y = 3.
Solution:
(i) Let A = \(\left[\begin{array}{rr}
0.8 & -0.6 \\
0.6 & 0.8
\end{array}\right]\);
and |A| = \(\left|\begin{array}{rr}
0.8 & -0.6 \\
0.6 & 0.8
\end{array}\right|\) = 0.64 + 0.36 = 1.0
∴ |A| = 1 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
The cofactors of R₁ are ; 0.8 ; -0.6
The cofactors of R₂ are ; 0.6 ; 0.8
∴ adj A = \(\left[\begin{array}{rr}
0.8 & -0.6 \\
0.6 & 0.8
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
0.8 & 0.6 \\
-0.6 & 0.8
\end{array}\right]\)
Thus, A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A = \(\left[\begin{array}{rr}
0.8 & 0.6 \\
-0.6 & 0.8
\end{array}\right]\)
Given system of eqn’s is equivalent to AX = B
where A = \(\left[\begin{array}{rr}
0.8 & -0.6 \\
0.6 & 0.8
\end{array}\right]\); X = \(\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right]\)
& B = \(\left[\begin{array}{l}
10 \\
20
\end{array}\right]\)
Since AX = B ⇒ X = A^-1B
⇒ \(\begin{aligned}
{\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right]=\left[\begin{array}{rr}
0.8 & 0.6 \\
-0.6 & 0.8
\end{array}\right]\left[\begin{array}{l}
10 \\
20
\end{array}\right] } & =\left[\begin{array}{c}
8+12 \\
-6+16
\end{array}\right]=\left[\begin{array}{l}
20 \\
10
\end{array}\right]
\end{aligned}\)
Thus, x₁ = 20 & ; x₂ = 10 be the required solution.

(ii) Let A = \(\left[\begin{array}{ll}
6 & 7 \\
4 & 5
\end{array}\right]\);
Here |A| = \(\left|\begin{array}{ll}
6 & 7 \\
4 & 5
\end{array}\right|\)
= 30 – 28 = 2 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
The cofactors of R₁ are ; 5 ; – 4
The cofactors of R₂ are ; – 7 ; 6
∴ adj A = \(\left[\begin{array}{rr}
5 & -4 \\
-7 & 6
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
5 & -7 \\
-4 & 6
\end{array}\right]\)
Thus, A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A = \(\frac{1}{2}\left[\begin{array}{rr}
5 & -7 \\
-4 & 6
\end{array}\right]\)
Given system of eqn’s is equivalent to AX = B
where A = \(\left[\begin{array}{ll}
6 & 7 \\
4 & 5
\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}
x \\
y
\end{array}\right] \& \mathrm{~B}=\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
⇒ X = A^-1B
⇒ X = \(\begin{array}{r}
{\left[\begin{array}{l}
x \\
y
\end{array}\right]=\frac{1}{2}\left[\begin{array}{rr}
5 & -7 \\
-4 & 6
\end{array}\right]\left[\begin{array}{l}
2 \\
3
\end{array}\right]} \\
=\frac{1}{2}\left[\begin{array}{l}
10-21 \\
-8+18
\end{array}\right]
\end{array}\)
⇒ \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
-11 / 2 \\
5
\end{array}\right]\)
∴ x = \(\frac { -11 }{ 2 }\); y = 5

Question 9.
(i) If A = \(\left[\begin{array}{rrr}
4 & -5 & -11 \\
1 & -3 & 1 \\
2 & 3 & -7
\end{array}\right]\), find A^-1, the system of equations
4x – 5y – 11z = 12, x – 3y + z = 1, 2x + 3y – 7z = 2.
(ii) If A = \(\left[\begin{array}{rrr}
8 & -4 & 1 \\
10 & 0 & 6 \\
8 & 1 & 6
\end{array}\right]\), find A^-1 solve the following system of linear equations : 8x – 4y + z = 5, 10x + 6z = 4, 8x + y + 6z = \(\frac { 5 }{ 2 }\).
Solution:
(i) Given A = \(\left[\begin{array}{rrr}
4 & -5 & -11 \\
1 & -3 & 1 \\
2 & 3 & -7
\end{array}\right]\);
Here |A| = \(\left|\begin{array}{rrr}
4 & -5 & -11 \\
1 & -3 & 1 \\
2 & 3 & -7
\end{array}\right|\);
Expanding along R₁.
= 4(+21 – 3) + 5 (-7 – 2) – 11(3 + 6)
= +72 – 45 – 99 = – 72 ≠ 0
∴ A^-1 exists & A^-1= \(\frac{1}{|\mathrm{~A}|}\) adj A … (1)
The cofactors of R₁ are ;

Given system of eqn’s is equivalent to AX = B
where X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) & B = \(\left[\begin{array}{c}
12 \\
1 \\
2
\end{array}\right]\)
Since |A| ≠ 0
∴ given system of eqns has unique solⁿ
⇒ X = A^-1B

Expanding along R₁.
= 8(0 – 6) + 4(60 – 48) + 1(10 – 0)
= – 48 + 48 + 10 = 10 ≠ 0
∴ A^-1 exists & A^-1= \(\frac{1}{|\mathrm{~A}|}\) adj A … (1)
The cofactors of R₁ are ;

The Given system of eqn’s is equivalent to AX = B
where X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) & B = \(\left[\begin{array}{c}
5 \\
4 \\
5 / 2
\end{array}\right]\)
Since |A| ≠ 0 ∴ given system of eqns has unique solution and is given by X = A^-1B

Question 10.
Find the product of two matrices A and B, where A = \(\left[\begin{array}{rrr}
-5 & 1 & 3 \\
7 & 1 & -5 \\
1 & -1 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
1 & 1 & 2 \\
3 & 2 & 1 \\
2 & 1 & 3
\end{array}\right]\) and use it to solve the system of equations x + y + 2z = 1, 3x + 2y + z = 7, 2x + y + 3z = 2.
Solution:
Here

The given system of eqn’s be
x + y + 2z = 1,
3x + 2y + z = 7,
2x + y + 3z = 2.
and this system is equivalent to BX = C
where X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) & C = \(\left[\begin{array}{l}
1 \\
7 \\
2
\end{array}\right]\)
Here |B| = \(\left|\begin{array}{lll}
1 & 1 & 2 \\
3 & 2 & 1 \\
2 & 1 & 3
\end{array}\right|\); Expanding along R₁.
= 1(6 – 1) – 1(9 – 2) + 2(3 – 4)
= 5 + 7 – 2 = – 4 ≠ 0 ∴ B^-1 exists
and given system of eqn’s has unique solution
Since BX = C ⇒ X = B^-1C

Use matrix method to examine the following systems of equations of consistency or inconsistency :

Question 11.
3x – 2y = 5
6x – 4y = 9
Solution:
Given system of equations is equivalent to AX = B
Where A = \(\left[\begin{array}{ll}
3 & -2 \\
6 & -4
\end{array}\right] \mathrm{X}=\left[\begin{array}{l}
x \\
y
\end{array}\right] \text { \& } \mathrm{B}\left[\begin{array}{l}
5 \\
9
\end{array}\right]\);
Here |A| = \(\left|\begin{array}{ll}
3 & -2 \\
6 & -4
\end{array}\right|\) = – 12 + 12 = 0
∴ A be singular matrix.
The cofactors of R₁ are ; – 4 ; – 6
The cofactors of R₂ are ; 2 ; 3
∴ adj A = \(\left[\begin{array}{rr}
-4 & -6 \\
2 & 3
\end{array}\right]^{\prime}=\left[\begin{array}{ll}
-4 & 2 \\
-6 & 3
\end{array}\right]\)
Now (adj A)B = \(\left[\begin{array}{ll}
-4 & 2 \\
-6 & 3
\end{array}\right]\left[\begin{array}{l}
5 \\
9
\end{array}\right]\)
= \(\left[\begin{array}{l}
-20+18 \\
-30+27
\end{array}\right]=\left[\begin{array}{l}
-2 \\
-3
\end{array}\right]\) ≠ 0
∴ given system of eqn’s has no solution. Thus the given system is inconsistent.

Question 12.
4x – 2y = 3
6x – 3y = 5
Solution:
The Given system of equations is equivalent to AX = B
Where A = \(\left[\begin{array}{ll}
4 & -2 \\
6 & -3
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) & B = \(\left[\begin{array}{l}
3 \\
5
\end{array}\right]\)
Here A = \(\left|\begin{array}{ll}
4 & -2 \\
6 & -3
\end{array}\right|\) = – 12 + 12 = 0
∴ A be singular matrix.
The cofactors of R₁ are ; – 3 ; – 6
The cofactors of R₂ are ; 2 ; 4
∴ adj A = \(\left[\begin{array}{rr}
-3 & -6 \\
2 & 4
\end{array}\right]^{\prime}=\left[\begin{array}{ll}
-3 & 2 \\
-6 & 4
\end{array}\right]\)
Now (adj A)B = \(\left[\begin{array}{ll}
-3 & 2 \\
-6 & 4
\end{array}\right]\left[\begin{array}{l}
3 \\
5
\end{array}\right]\)
= \(\left[\begin{array}{c}
-9+10 \\
-18+20
\end{array}\right]=\left[\begin{array}{l}
1 \\
2
\end{array}\right]\) ≠ 0
The given system has no solution and given system is inconsistent.

Question 13.
x + 5y = 3
2x + 10y = 6
Solution:
The Given system of equations is equivalent to AX = B
Where A = \(\left[\begin{array}{cc}
1 & 5 \\
2 & 10
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) & B = \(\left[\begin{array}{l}
3 \\
6
\end{array}\right]\)
Here A = \(\left|\begin{array}{cc}
1 & 5 \\
2 & 10
\end{array}\right|\) = 10 – 10 = 0
∴ A be singular.
The cofactors of R₁ are ; 10 ; – 2
The cofactors of R₂ are ; -5 ; 1
∴ adj A = \(\left[\begin{array}{rr}
10 & -2 \\
-5 & 1
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
10 & -5 \\
-2 & 1
\end{array}\right]\)
Here (adj A)B = \(\left[\begin{array}{cc}
10 & -5 \\
-2 & 1
\end{array}\right]\left[\begin{array}{l}
3 \\
6
\end{array}\right]\)
= \(\left[\begin{array}{c}
30-30 \\
-6+6
\end{array}\right]=\left[\begin{array}{l}
0 \\
0
\end{array}\right]\) = 0
The given system has infinitly many solutions and is consistent. Let y = k and putting y = k in first eqn.
x + 5k = 3 ⇒ x = 3 – 5 k
putting x = 3 – 5k and y = k in second eqn., we have
2(3 – 5k) + 10k = 6
⇒ 6 – 10k + 10k = 6
⇒ 6 = 6, which is true.
Thus, given system of equations is consistent and has infinitly many solutions given by x = 3 – 5x; y = k ∈ R

Question 14.
3x – y + 2z = 3
2x + y + 3z = 5
x – 2y – z = 1
Solution:
The Given system of equations is equivalent to AX = B
where A = \(\left[\begin{array}{ccc}
3 & -1 & 2 \\
2 & 1 & 3 \\
1 & -2 & -1
\end{array}\right]\) & X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) & B = \(\left[\begin{array}{l}
3 \\
5 \\
1
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ccc}
3 & -1 & 2 \\
2 & 1 & 3 \\
1 & -2 & -1
\end{array}\right|\)
Expanding along R₁.
= 3(- 1 + 6) + 1(- 2 – 3) + 2(- 4 – 1)
= 15 – 5 – 10 = 0
Thus, the given system of eqns has either no solution or infinitly many solution.
The cofactors of R₁ are ;
\(\left|\begin{array}{cc}
1 & 3 \\
-2 & -1
\end{array}\right| ;-\left|\begin{array}{cc}
2 & 3 \\
1 & -1
\end{array}\right| ;\left|\begin{array}{cc}
2 & 1 \\
1 & -2
\end{array}\right|\)
i.e. 5 ; 5 ; -5
The cofactors of R₂ are ;
\(-\left|\begin{array}{cc}
-1 & 2 \\
-2 & -1
\end{array}\right| ;\left|\begin{array}{cc}
3 & 2 \\
1 & -1
\end{array}\right| ;-\left|\begin{array}{cc}
3 & -1 \\
1 & -2
\end{array}\right|\)
i.e. -5 ; -5 ; 5
The cofactors of R₃ are ;

The given system of eqn’s has no solution and system is inconsistent.

Question 15.
x + y + z = 6
x + 2y + 3z = 14
x + 4y + Iz = 30
Solution:
The Given system of equations is equivalent to AX = B
where X = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 4 & 7
\end{array}\right]\) & X = \(=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) & B = \(\left[\begin{array}{c}
6 \\
14 \\
30
\end{array}\right]\)
Here A = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 4 & 7
\end{array}\right|\);
Expanding along R₁.
= 1(14 – 12) -1(7 – 3) + 1(4 – 2)
= 2 – 4 + 2 = 0
∴ A be singular matrix.
Thus, the given system of eqn’s has either no solution or infinitly many solution.
The cofactors of R₁ are ;

The given system of eqn’s is consistent and has infinitly many solution.
Putting z = k in first two eqn. of given system, we get
x + y = 6 – k …(1)
x + 2y = 14 – 3k …(2)
on solving (1) & (2); we have
y = 14 – 3k – 6 + k = 8 – 2k & x = k – 2 and the values of x, y and z satisfies the 3rd eqn. of given system of eqns.
Thus, x = k – 2 ;y = 8 – 2k ; z = k,
where K ∈ R be the required infinitely many solution.

Question 16.
2x – y + 3z = 1
x + 2y – z = 2
5y – 5z = 3
Solution:
The Given system of equations is equivalent to AX = B
where X = \(\left[\begin{array}{ccc}
2 & -1 & 3 \\
1 & 2 & -1 \\
0 & 5 & -5
\end{array}\right]\) & X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) & B = \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
Here A = \(\left|\begin{array}{ccc}
2 & -1 & 3 \\
1 & 2 & -1 \\
0 & 5 & -5
\end{array}\right|\);
Expanding along R₁.
= 1(-10 + 5) + 1(-5 – 0) + 3(5 – 0)
= – 10 – 5 + 15 = 0
A be singular matrix.
Thus, the given system of eqn’s has either no solution or infinitly many solution.
The cofactors of R₁ are ;

Thus given system of eqn’s is consistent and have infinitly may solution.
Putting z = k in first two eqn. of given system, we have
2x – y = 1 – 3k …(1)
x + 2y = 2 + k …(2)
Multiplying eqn. (1) by 2 + eqn. (2); we get
5x = 2 – 6k + 2 + k ⇒ x = \(\frac{4-5 k}{5}\)
∴ from (1) ; we have
y = \(\frac{8-10 k}{5}-1+3 k=\frac{3+5 k}{5}\)
Putting x = \(\frac{4-5 k}{5}\) , y = \(\frac{3+5 k}{5}\), z = k in third eqn. 5y – 5z =3
i.e. \(\left(\frac{3+5 k}{5}\right)\) – 5k = 3 ⇒ 3 = 3, which is true.
Hence the given system of eqn’s has infinite many solutions and is given by
x = \(\frac{4-5 k}{5}\) , y = \(\frac{3+5 k}{5}\) & z = k
where k ∈ R.

Question 17.
Show that the following system of equations is consistent x-2y + z = 0, y-z = 3, 2x-3z = 10 Also, find the solution using matrix method.
Solution:
The given system of eqns is equivalent to AX = B where A = \(\left[\begin{array}{rrr}
1 & -2 & 1 \\
0 & 1 & -1 \\
2 & 0 & -3
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
y
\end{array}\right]\) & B = \(\left[\begin{array}{c}
0 \\
3 \\
10
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
1 & -2 & 1 \\
0 & 1 & -1 \\
2 & 0 & -3
\end{array}\right|\);
Expanding along R₁
= 1 (- 3 – 0) + 2(0 + 2) + 1(0 – 2)
= – 3 + 4 – 2 = – 1 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adjA
Thus given system of eqn’s has unique solution.
The cofactors of R₁ are :

Question 18.
Find k so that the equations
3x – 2y + 2z = 1, 2x + y + 3z = – 1, x – 3y + kz = 0 may have a unique solution.
Solution:
Given system of equations is equivalent to AX = B where A = \(\left[\begin{array}{rrr}
3 & -2 & 2 \\
2 & 1 & 3 \\
1 & -3 & k
\end{array}\right]\) & X = \(\left[\begin{array}{l}
x \\
y \\
y
\end{array}\right]\) & B = \(\left[\begin{array}{c}
1 \\
-1 \\
0
\end{array}\right]\)
Since given system of eqn’s has unique solution.
∴ |A| ≠ 0 ⇒ \(\left|\begin{array}{rrr}
3 & -2 & 2 \\
2 & 1 & 3 \\
1 & -3 & k
\end{array}\right|\) ≠ 0;
Expanding along R₁.
⇒ 3(k + 9) + 2(2k – 3) + 2(- 6 – 1) ≠ 0
⇒ 3k + 27 + 4k – 6 – 14 ≠ 0
⇒ 7k + 7 ≠ 0
⇒ k ≠ – 1

Question 19.
For what value off k, do the equations
2x – 3y + 2z = a
5x + 4y – 2z = – 3
x – 13y + kz = 9
not have a unique solution.
Solution:
The given system of equations is equivalent to AX = B where A = \(\left[\begin{array}{rrr}
2 & -3 & 2 \\
5 & 4 & -2 \\
1 & -13 & k
\end{array}\right]\) & X = \(\left[\begin{array}{l}
x \\
y \\
y
\end{array}\right]\) & B = \(\left[\begin{array}{c}
a \\
-3 \\
9
\end{array}\right]\)
Here, |A| = \(\left|\begin{array}{rrr}
2 & -3 & 2 \\
5 & 4 & -2 \\
1 & -13 & k
\end{array}\right|\)
Expanding along R₁
= 2(4k – 26) + 3(5k + 2) + 2(- 65 – 4)
= 8k – 52 + 15k + 6 – 138
= 23k – 184
Since the given system do not have unique solution
∴ |A| = 0
⇒ 23k = 184
⇒ k = 8

Question 20.
Suppose the demand curve for automobiles over some time period can be written x₁ = 15000 – 0.2x₂ where x₁ is the price of an automobile and x₂ is the correponding quamtity. Suppose the supply curve is x₁ = 600 + 0.4 x₂. Use matrix theory to obtain x₁.
Solution:
Given system of equations be,
x₁ = 15000 – 0.2x₂
⇒ x₁ + 0.2x₂ = 15000
x₁ = 600 + 0.4 x₂
⇒ x₁ – 0.4 x₂ = 600
The given system of equations is equivalent to AX = B where
A = \(\left[\begin{array}{cc}
1 & 0.2 \\
1 & -0.4
\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right] \& \mathrm{~B}=\left[\begin{array}{c}
15000 \\
600
\end{array}\right]\)
Here, |A| = \(\left|\begin{array}{cc}
1 & 0.2 \\
1 & -0.4
\end{array}\right|\) = – 0.4 – 0.2 = – 0.6 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
and the given system of eqn’s has unique solution.
The cofactors of R₁ are : -0.4; -1
The cofactors of R₂ are : -0.2; 1

Question 21.
Gaurav purchases 3 pens, 2 bags and 1 instrument box and pays ₹ 41. From the same shop, Dheeraj purchases 2 pens, 1 bag and 2 instrument boxes and pays ₹ 29, while Ankur purchases 2 pens, 2 bags and 2 instrument boxes and pays 44. Translate the problem into a system of equations. Solve the system of equation by matrix method and hence find the cost of one pen one bag and one instrument box.
Solution:
Let the cost of one pen be Rs x and of one bag be Rs y and of one instrument box be Rs z.
Thus the mathematical modiling of given problem be given as under :
3x + 2y + z = 41
2x + y + 2z = 29
2x + 2y + 2z = 44
Given system of eqns is equivalent to
AX = B where A = \(\left[\begin{array}{lll}
3 & 2 & 1 \\
2 & 1 & 2 \\
2 & 2 & 2
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) & B = \(\left[\begin{array}{l}
41 \\
29 \\
44
\end{array}\right]\)
Expanding along R₁
= 3(2 – 4) – 2(4 – 4) + 1(4 – 2)
= – 6 – 0 + 2 = – 4 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adjA
Also, given system of eqn’s has unique solution.
The cofactors of R₁ are ;

Hence the required cost of each pen be Rs 2, each bag be Rs 15 and cost of each instrument be Rs. 5

Question 22.
A school wants to award its students for the value of Honesty, Regularity and Hard work with a total cash award of ₹ 6000. Three times the award money for Hardwork added to that given for Honesty amounts to ₹ 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically, and find the award money for each value, using matrix method.
Solution:
Let the award money for honesty. Regurity and hardwork be ₹ x, ₹ y and ₹ z respectively.
The mathematical modelling of given problem is given as under :
x + y + z = 6000
x + 3z = 11,000
x + z = 2y
Given system of eqns is equivalent to
AX = B where A = \(\left[\begin{array}{rrr}
1 & 1 & 1 \\
1 & 0 & 3 \\
1 & -2 & 1
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) & B = \(\left[\begin{array}{c}
6000 \\
11000 \\
0
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
1 & 1 & 1 \\
1 & 0 & 3 \\
1 & -2 & 1
\end{array}\right|\);
Expanding along R₁
= 1(0 + 6) – 1(1 – 3) + 1(- 2 – 0)
= 6 + 2 – 2 = 6 ≠ 0
∴ A^-1 exists and given system of eqn’s has unique solution.
The cofactors of R₁ are :

Thus, x = 500 ; y = 2000 ; z = 3500
Hence the award money given by honesty, regularity and hard work be ₹ 500, ₹ 2000 & ₹ 3500 respectively.

Examples

Question 1.
Using matrices, solve the following system of equations :
x₁ – 2x₂ + 3x₃ = 4
2x₁ + x₂ – 3x₃ = 5
– x₁ + x₂ + 2x₃ = 3
Solution:
Given system of eqn’s is equivalent to AX =B
where A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
2 & 1 & -3 \\
-1 & 1 & 2
\end{array}\right]\); X = \(\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]\); B = \(\left[\begin{array}{l}
4 \\
5 \\
3
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
1 & -2 & 3 \\
2 & 1 & -3 \\
-1 & 1 & 2
\end{array}\right|\);
Expanding along R₁
= 1(2 + 3) + 2(4 – 3) + 3(2 + 1)
= 5 + 2 + 9 = 16 ≠ 0
∴ A^-1 exists
and system of eqns has unique solution given by X = A^-1 B
The cofactors of R₁ are :
\(\left|\begin{array}{rr}
1 & -3 \\
1 & 2
\end{array}\right| ;-\left|\begin{array}{rr}
2 & -3 \\
-1 & 2
\end{array}\right| ;\left|\begin{array}{rr}
2 & 1 \\
-1 & 1
\end{array}\right|\)
i.e. 5; – 1 ; 3
The cofactors of R₂ are ;
\(-\left|\begin{array}{rr}
-2 & 3 \\
1 & 2
\end{array}\right| ;\left|\begin{array}{rr}
1 & 3 \\
-1 & 2
\end{array}\right| ;-\left|\begin{array}{rr}
1 & -2 \\
-1 & 1
\end{array}\right|\)
i.e. 7; 5; +1
The cofactors of R₃ are :
\(\left|\begin{array}{rr}
-2 & 3 \\
1 & -3
\end{array}\right| ;-\left|\begin{array}{rr}
1 & 3 \\
2 & -3
\end{array}\right| ;\left|\begin{array}{rr}
1 & -2 \\
2 & 1
\end{array}\right|\)
i.e. 3; 9; 5

Question 2.
Using matrix method, solve the following system of equations :
x – 2y + 3z = 6
x + 4y + z = 12
x – 3y + 2z = 1
Solution:
The given system of eqn’s is equivalent
where A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
1 & 4 & 1 \\
1 & -3 & 2
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y \\
y
\end{array}\right]\); B = \(\left[\begin{array}{c}
6 \\
12 \\
1
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
1 & -2 & 3 \\
1 & 4 & 1 \\
1 & -3 & 2
\end{array}\right|\);
Expanding along R₁
= 1(8 + 3) + 2(2 – 1) + 3(-3 – 4)
= 11 + 2 – 21 = – 8 ≠ 0
∴ A^-1 exists
Thus given system of eqn’s has unique solution given by X = A^-1 B
The cofactors of R₁ are :

Thus, x = 1, y = 2 ; z = 3

Question 3.
Find the inverse of the matrix A = \(\left[\begin{array}{ll}
3 & 1 \\
4 & 2
\end{array}\right]\).
Solution:
Given A = \(\left[\begin{array}{ll}
3 & 1 \\
4 & 2
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ll}
3 & 1 \\
4 & 2
\end{array}\right|\) = 6 – 4 = 2 ≠ 0
∴ A^-1 exists
The cofactors of R₁ are : 2, – 4
The cofactors of R₂ are : – 1, 3
∴ adj A = \(\left[\begin{array}{cc}
2 & -4 \\
-1 & 3
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
2 & -1 \\
-4 & 3
\end{array}\right]\)
Thus, A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A = \(\frac{1}{2}\left[\begin{array}{cc}
2 & -1 \\
-4 & 3
\end{array}\right]\)

Question 4.
If \(\left[\begin{array}{lll}
x^2 & 3 & 4 \\
1 & 9 & 8
\end{array}\right]+\left[\begin{array}{crr}
-3 x & 1 & -5 \\
-3 & -2 & -6
\end{array}\right]\) = \(\left[\begin{array}{rrr}
4 & 4 & -1 \\
-2 & 7 & 2
\end{array}\right]\), find the value sof x.
Solution:
Given \(\left[\begin{array}{lll}
x^2 & 3 & 4 \\
1 & 9 & 8
\end{array}\right]+\left[\begin{array}{crr}
-3 x & 1 & -5 \\
-3 & -2 & -6
\end{array}\right]\) = \(\left[\begin{array}{rrr}
4 & 4 & -1 \\
-2 & 7 & 2
\end{array}\right]\)
⇒ \(\left[\begin{array}{ccc}
x^2-3 x & 3+1 & 4-5 \\
1-3 & 9-2 & 8-6
\end{array}\right]=\left[\begin{array}{ccc}
4 & 4 & -1 \\
-2 & 7 & 2
\end{array}\right]\)
⇒ \(\left[\begin{array}{ccc}
x^2-3 x & 4 & -1 \\
-2 & 7 & +2
\end{array}\right]=\left[\begin{array}{ccc}
4 & 4 & -1 \\
-2 & 7 & 2
\end{array}\right]\)
Thus their corresponding entries are equal.
∴ x² – 3x = 4
⇒ x² – 3x – 4 = 0
⇒ (x + 1)(x – 4) = 0
⇒ x = – 1, 4

Question 5.
Using matrices, solve the following system of equations :
x + 2y = 5, y + 2z = 8, 2x + z = 5
Solution:
The given system of eqn’s is equivalent to AX = B
where A = \(\left[\begin{array}{lll}
1 & 2 & 0 \\
0 & 1 & 2 \\
2 & 0 & 1
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) & B = \(\left[\begin{array}{l}
5 \\
8 \\
5
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{lll}
1 & 2 & 0 \\
0 & 1 & 2 \\
2 & 0 & 1
\end{array}\right|\);
Expanding along R₁
= 1(1 – 0) – 2(0 – 4) + 0
∴ |A| = 1 + 8 = 9 ≠ 0
∴ A^-1 exists given system of eqn’s has unique solution given by X = A^-1 B
The cofactors of R₁ are :

Thus, x = 1 ; y = 2 ; z = 3

Question 6.
If the matrix A = \(\left[\begin{array}{rr}
2 & 3 \\
5 & -2
\end{array}\right]\), show that A^-1 = \(\frac { 1 }{ 19 }\)A.
Solution:
Given A = \(\left[\begin{array}{rr}
2 & 3 \\
5 & -2
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{cc}
2 & 3 \\
5 & -2
\end{array}\right|\) = – 4 – 15 = – 19 ≠ 0
Thus A^-1 exists and A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A … (1)
The cofactors of R₁ are : -2 ; -5
The cofactors of R₂ are : -3 ; 2
∴ adj A = \(\left[\begin{array}{cc}
-2 & -5 \\
-3 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
-2 & -3 \\
-5 & 2
\end{array}\right]\)
∴ from (1); A^-1 = – \(\frac{1}{19}\left[\begin{array}{cc}
-2 & -3 \\
-5 & 2
\end{array}\right]\)
= \(\frac{1}{19}\left[\begin{array}{cc}
2 & 3 \\
5 & -2
\end{array}\right]=\frac{1}{19}\)A

Question 7.
Given that A = \(\left[\begin{array}{rr}
\cos x & \sin x \\
-\sin x & \cos x
\end{array}\right]\) and A (adj A) = k\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), find the value of k.
Solution:

Question 8.
Given the matrix A = \(\left[\begin{array}{rrr}
1 & 0 & 2 \\
-2 & 1 & 0 \\
0 & -1 & 2
\end{array}\right]\), compute A^-1.
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & 0 & 2 \\
-2 & 1 & 0 \\
0 & -1 & 2
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
1 & 0 & 2 \\
-2 & 1 & 0 \\
0 & -1 & 2
\end{array}\right|\);
Expanding along R₁
= 1(2 – 0) + 0 + 2(2 – 0)
= 2 + 4 ≠ 0
∴ A^-1 exists and A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A … (1)
The cofactors of R₁ are :

Question 9.
Find the adjoint of the matrix A = \(\left[\begin{array}{rrr}
1 & 0 & -1 \\
3 & 4 & 5 \\
0 & -6 & -7
\end{array}\right]\) and hrence find the matrix A^-1.
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & 0 & -1 \\
3 & 4 & 5 \\
0 & -6 & -7
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
1 & 0 & -1 \\
3 & 4 & 5 \\
0 & -6 & -7
\end{array}\right|\);
Expanding along R₁
= 1(- 28 + 30) + 0 – 1(- 18 – 0)
= 2 + 18 = 20 ≠ 0
∴ A^-1 exists and A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A … (1)

Question 10.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\), show that A² – 3I = 2A.
Solution:
Given A = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\)
L.H.S = A² – 3I
= \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]-3\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1+4 & 2+2 \\
2+2 & 4+1
\end{array}\right]-\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\)
= \(\left[\begin{array}{ll}
5-3 & 4-0 \\
4-0 & 5-3
\end{array}\right]=\left[\begin{array}{ll}
2 & 4 \\
4 & 2
\end{array}\right]\)
= 2\(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\) = 2A
= R.H.S
Thus, A² – 3I = 2A

Question 11.
Solve the following system of equations using matrices :
x + y + z = 6, x – y + z = 2,
Solution:
The given system of eqns is equivalent to AX = B where A = \(\left[\begin{array}{rrr}
1 & 1 & 1 \\
1 & -1 & 1 \\
2 & 1 & -1
\end{array}\right]\); X = \(\left[\begin{array}{l}
x \\
y \\
y
\end{array}\right]\); B = \(\left[\begin{array}{l}
6 \\
2 \\
1
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
1 & 1 & 1 \\
1 & -1 & 1 \\
2 & 1 & -1
\end{array}\right|\);
Expanding along R₁
|A| = 1(1 – 1) – 1(- 1 – 2) + 1(1 + 2)
= 0 + 3 + 3 ≠ 0
∴ A^-1 exists and given system of eqn’s has unique solution given by X = A^-1 B
The cofactors of R₁ are :

Question 12.
Find x and y, if x + y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) and x – y = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\).
Solution:

Question 13.
If A = \(\left[\begin{array}{rrr}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right]\), find A^-1 and hence solve the following system of linear equations:
x + 2y – 3z = – 4, 2x + 3y + 2z = 2, 3x – 3y – 4z = 11.
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right|\);
Expanding along R₁
= 1(- 12 + 6) – 2(- 8 – 6) – 3(- 6 – 9)
= – 6 + 28 + 45 = 67 ≠ 0
∴ A^-1 exists and A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A … (1)
The cofactors of R₁ are :

Thus, x = 3 ; y = – 2 & z = 1 be the required solution.

Question 14.
If M (θ) = \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), show that M(x) M(y) = M(x + y)
Solution:

Question 15.
Solve the following linear equations using the matrix method :
x + y + z = 9,
2x + 5y + 7z = 52,
2x + y – z = 0.
Solution:
The given system of eqns is equivalent to AX = B where A = \(\left[\begin{array}{rrr}
1 & 1 & 1 \\
2 & 5 & 7 \\
2 & 1 & -1
\end{array}\right]\) & X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) ; B = \(\left[\begin{array}{c}
9 \\
52 \\
0
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
1 & 1 & 1 \\
2 & 5 & 7 \\
2 & 1 & -1
\end{array}\right|\);
Expanding along R₁
|A| = 1(- 5 – 7) – 1(- 2 – 14) + 1(2 – 10)
= – 12 + 16 – 8 = – 4 ≠ 0
∴ A^-1 exists and given system of eqn’s has unique solution given by X = A^-1 B
The cofactors of R₁ are :

Thus, x = 1; y = 3 ; z = 5 be the required solution

Question 16.
If the matrix A = \(\left[\begin{array}{rrr}
6 & x & 2 \\
3 & -1 & 2 \\
-10 & 5 & 2
\end{array}\right]\) is a singular matrix, find the value of x.
Solution:
Given A = \(\left[\begin{array}{rrr}
6 & x & 2 \\
2 & -1 & 2 \\
-10 & 5 & 2
\end{array}\right]\)
Since A be a singular matrix.
∴ |A| = 0
i.e., \(\left|\begin{array}{rrr}
6 & x & 2 \\
2 & -1 & 2 \\
-10 & 5 & 2
\end{array}\right|\) = 0
Expanding along R₁;
we have 6(- 2 – 10) – x(4 + 20) + 2(10 – 10) = 0 ⇒ – 72 – 24x = 0 ⇒ x = – 3

Question 17.
Using matrix method, solve the following equations:
5x + 3y + z = 16, 2x + y + 3z = 19, x + 2y + 4z = 25.
Solution:
The given system of eqns is equivalent to AX = B where A = \(\left[\begin{array}{lll}
5 & 3 & 1 \\
2 & 1 & 3 \\
1 & 2 & 4
\end{array}\right]\) & X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) ; B = \(\left[\begin{array}{l}
16 \\
19 \\
25
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{lll}
5 & 3 & 1 \\
2 & 1 & 3 \\
1 & 2 & 4
\end{array}\right|\); Expanding along R₁
= 5(4 – 6) – 3(8 – 3) + 1(4 – 1)
= – 10 – 15 + 3 = – 22 ≠ 0
∴ A^-1 exists and given system of eqn’s has unique solution given by X = A^-1 B … (1)
The cofactors of R₁ are :

Thus, x = 1 ; y = 2 ; z = 5 be the required solution

Question 18.
If A = \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\), find x such that A² = xA – 2I. Hence, find A^-1.
Solution:
Given A = \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right|\) = – 6 + 8 = 2 ≠ 0
∴ A^-1 exists
Since A² = xA – 2I … (1)
\(\left[\begin{array}{ll}
3 & –2 \\
4 & -2
\end{array}\right]\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]=x\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]-2\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
9-8 & -6+4 \\
12-8 & -8+4
\end{array}\right]=\left[\begin{array}{cc}
3 x & -2 x \\
4 x & -2 x
\end{array}\right]-\left[\begin{array}{cc}
2 & 0 \\
0 & 2
\end{array}\right]\)
\(\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]=\left[\begin{array}{cc}
3 x-2 & -2 x \\
4 x & -2 x-2
\end{array}\right]\)
Thus corresponding elements
∴ 1 = 3x – 2 ⇒ 3x = 3 ⇒ x = 1
-2x = – 2 ⇒ x = 1
Also, 4x = 4 ⇒ x = 1
and – 4 = – 2x – 2 ⇒ x = 1
Hence, x = 1
∴ from (1) ; A² = A – 2I
2I = A – A² … (2)
pre multiplying both sides of eqn (2) by A^-1 ; we have

Question 19.
Solve the following system of equations using matrix method :
\(\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4, \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1,\), \(\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2\)
Solution:
Putting \(\frac { 1 }{ x }\) = u; \(\frac { 1 }{ y }\) = v & \(\frac { 1 }{ z }\) = w in
given system of eqn’s ; we have
2u + 3v + 10w = 4
4u + 6v + 5w = 1
6u + 9v – 20w = 2
The given system of eqn’s is eqrualant to AX = B
where A = \(\left[\begin{array}{ccc}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right]\) & X = \(\left[\begin{array}{l}
u \\
v \\
w
\end{array}\right]\) ; B = \(\left[\begin{array}{l}
4 \\
1 \\
2
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ccc}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right|\)
Taking 2, 3 & 5 common from R₁; R₂ & R₃ respectively
= 2 x 3 x 5\(\left|\begin{array}{ccc}
1 & 1 & 2 \\
2 & -2 & 1 \\
3 & 3 & -4
\end{array}\right|\) ;
Expanding along R₁
= 30[1(8 – 3) – 1(- 8 – 3) +2(6 + 6)]
= 30[5 + 11 + 24]
= 30 x 40
= 1200 ≠ 0
∴ A^-1 exists and A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A … (1)
Thus, given system has unique solution.
The cofactors of R₁ are ;

Question 20.
Solve for x and y if
\(\left[\begin{array}{l}
x^2 \\
y^2
\end{array}\right]+2\left[\begin{array}{l}
2 x \\
3 y
\end{array}\right]=3\left[\begin{array}{c}
7 \\
-3
\end{array}\right]\).
Solution:
Given \(\left[\begin{array}{l}
x^2 \\
y^2
\end{array}\right]+2\left[\begin{array}{l}
2 x \\
3 y
\end{array}\right]=3\left[\begin{array}{c}
7 \\
-3
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x^2 \\
y^2
\end{array}\right]+\left[\begin{array}{l}
4 x \\
6 y
\end{array}\right]=\left[\begin{array}{c}
21 \\
-9
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x^2+4 x \\
y^2+6 y
\end{array}\right]=\left[\begin{array}{l}
21 \\
-9
\end{array}\right]\)
Thus corresponding entries are equal.
∴ x² + 4x – 21 = 0
⇒ (x – 3) (x + 7) = 0
⇒ x = 3, – 7
& y² + 6y = – 9
⇒ y² + 6y + 9 = 0
⇒ (y + 3)² = 0
⇒ y = – 3

Question 21.
Find the product of the matrices A and B, where
A = \(\left[\begin{array}{rrr}
-5 & 1 & 3 \\
7 & 1 & -5 \\
1 & -1 & 1
\end{array}\right]\), B = \(\left[\begin{array}{lll}
1 & 1 & 2 \\
3 & 2 & 1 \\
2 & 1 & 3
\end{array}\right]\).
Hence, solve the following equations by matrix method :
x + y + 2z = 1, 3x + 2y + z = 7, 2x + y + 3z = 2.
Solution:

The given system of eqn’s be
x + y + 2z = 1,
3x + 2y + z = 7,
2x + y + 3z = 2.
and this system is equivalent to BX = C
where X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) & C = \(\left[\begin{array}{l}
1 \\
7 \\
2
\end{array}\right]\)
Here |B| = \(\left|\begin{array}{lll}
1 & 1 & 2 \\
3 & 2 & 1 \\
2 & 1 & 3
\end{array}\right|\); Expanding along R₁
= 1(6 – 1) -1(9 – 2) +2(3- 4)
= 5 + 7 – 2 = – 4 ≠ 0
∴ B^-1 exists
and given system of eqn’s has unique solution
Since BX = C ⇒ X = B^-1C
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{4}\left[\begin{array}{ccc}
-5 & 1 & 3 \\
7 & 1 & -5 \\
1 & -1 & 1
\end{array}\right]\left[\begin{array}{l}
1 \\
7 \\
2
\end{array}\right]\)
= \(\frac{1}{4}\left[\begin{array}{c}
-5+7+6 \\
7+7-10 \\
1-7+2
\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}
8 \\
4 \\
-4
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
2 \\
1 \\
-1
\end{array}\right]\)
∴ x = 2 ; y = 1 and z = – 1

Question 22.
If (A – 2I) (A – 3I) = 0, when A = \(\left(\begin{array}{rr}
4 & 2 \\
-1 & x
\end{array}\right)\) and I = \(\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)\), find the value of x.
Solution:
Given A = \(\left(\begin{array}{rr}
4 & 2 \\
-1 & x
\end{array}\right)\) and I = \(\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)\)

Thus their corresponding entries are equal.
2x – 2 = 0
⇒ x = 1; 1 – x = 0 ⇒ x = 1
x² – 5x + 4 = 0 ⇒ x = 1, 4
Hence, the common value of x be 1

Question 23.
Find A^-1, where A = \(\left[\begin{array}{rrr}
4 & 2 & 3 \\
1 & 1 & 1 \\
3 & 1 & -2
\end{array}\right]\).
Hence, solve the following system of linear equation :
4x + 2y + 3z = 2
x + y + z = 1
3x + y – 2z = 5
Solution:
Given A = \(\left[\begin{array}{rrr}
4 & 2 & 3 \\
1 & 1 & 1 \\
3 & 1 & -2
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
4 & 2 & 3 \\
1 & 1 & 1 \\
3 & 1 & -2
\end{array}\right|\);
Expanding along R₁
= 4(-2 – 1) – 2(-2 – 3) + 3(1 – 3)
= -12 + 10 – 6 = – 18 ≠ 0
∴ A^-1 exists and given system of eqn’s has unique solution

Question 24.
If A = \(\left[\begin{array}{ll}
3 & 1 \\
7 & 5
\end{array}\right]\), find the values of x and y such that A² + xI₂ = yA.
Solution:
Given A = \(\left[\begin{array}{ll}
3 & 1 \\
7 & 5
\end{array}\right]\)
Also A² + xI₂ = yA
⇒ \(\left[\begin{array}{ll}
3 & 1 \\
7 & 5
\end{array}\right]\left[\begin{array}{ll}
3 & 1 \\
7 & 5
\end{array}\right]+x\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=y\left[\begin{array}{ll}
3 & 1 \\
7 & 5
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
9+7 & 3+5 \\
21+35 & 7+25
\end{array}\right]+\left[\begin{array}{ll}
x & 0 \\
0 & x
\end{array}\right]=\left[\begin{array}{cc}
3 y & y \\
7 y & 5 y
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
16+x & 8 \\
56 & 32+x
\end{array}\right]=\left[\begin{array}{cc}
3 y & y \\
7 y & 5 y
\end{array}\right]\)
Thus their corresponding entries are equal
∴ 16 + x = 3y … (1) ; y = 8
∴ from (1); 16 + x = 24 ⇒ x = 8
Also 7y = 56 ⇒ y = 8
& 32 + x = 5y ⇒ 5y = 40 ⇒ y = 8
Thus x = 8 & y = 8

Question 25.
Using matrix method, solve the following system of equations :
x – 2y = 10, 2x + y + 3z = 8 and – 2y + z = 7.
Solution:
The given system of eqns is equivalent to AX = B where A = \(\left[\begin{array}{rrr}
1 & -2 & 0 \\
2 & 1 & 3 \\
0 & -2 & 1
\end{array}\right]\)
& X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\); B = \(\left[\begin{array}{c}
10 \\
8 \\
7
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
1 & -2 & 0 \\
2 & 1 & 3 \\
0 & -2 & 1
\end{array}\right|\);
Expanding along R₁
= 1(1 + 6) + 2(2 – 0) + 0 = 11 ≠ 0
A^-1 exists and given system of eqn’s has unique solution given by X = A^-1 B … (1)
The cofactors of R₁ are :
Thus, x = 4 ; y = – 3 ; z = 1 be the required solution.

Question 26.
Find the value of k if M = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right]\) and M² – AM – I₂ = 0.
Solution:
Given M = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right]\)

Thus corresponding entries are equal.
∴ 5 – k = 0 ⇒ k = 4
8 – 2k = 0 ⇒ k = 4
Also k = 4 satisfies 13 – 3k – 1 = 0
Thus required value of k = 4

Question 27.
Given two matrices A and B
A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
1 & 4 & 1 \\
1 & -3 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
11 & -5 & -14 \\
-1 & -1 & 2 \\
-7 & 1 & 6
\end{array}\right]\), find AB and use this result to solve the following system of equations :
x – 2y + 3z = 6, x + 4y + z = 12, x – 3y + 2z = 1.
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
1 & 4 & 1 \\
1 & -3 & 2
\end{array}\right]\)

Expanding along R₁; we have
= 1(8 + 3) + 2(2 – 1) + 3(- 3 – 4)
= 11 + 2 – 21 = – 8 ≠ 0
∴ A^-1 exists and A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A … (1)
Since AB = – 8I₃ ; pre-multiplying both sides by A^-1 ; we have

Thus, x = 1; y = 2 & z = 3 be the required solution.

Question 28.
Find the matrix x for which
\(\left[\begin{array}{ll}
5 & 4 \\
1 & 1
\end{array}\right] X=\left[\begin{array}{rr}
1 & -2 \\
1 & 3
\end{array}\right]\).
Solution:

Question 29.
Solve the following system of linear equations using matrix method :
3x + y + z = 1, 2x + 2z = 0, 5x + y + 2z = 2
Solution:
The given system of eqns is equivalent to AX = B; where A = \(\left[\begin{array}{rrr}
1 & -2 & 0 \\
2 & 1 & 3 \\
0 & -2 & 1
\end{array}\right]\)
& X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\); B = \(\left[\begin{array}{c}
10 \\
8 \\
7
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
1 & -2 & 0 \\
2 & 1 & 3 \\
0 & -2 & 1
\end{array}\right|\);
Expanding along R₁ we have
= 3(0 – 2) – 1(4 – 10) + 1(2 – 0)
= – 6 + 6 + 2 = 2 ≠ 0
∴ A^-1 exists and given system of eqn’s has unique solution
The cofactors of R₁ are :

Question 30.
For a 3 x 3 matrix A = [a_ij] where elements are given by a_ij = \(\frac{|i-j|}{2}\) then the value of the element a₂₃ is ……………….
Solution:
Given a_ij = \(\frac{|i-j|}{2}\)
∴ a₂₃ = \(\frac{|2-3|}{2}=\frac{1}{2}\)

Question 31.
(i) If a matrix has 5 elements, then all possible orders it can have are ………………
(ii) The product of m x p and p x n matrix is an ……………… matrix.
Solution:
(i) We know that, a matrix of order m x n have mn elements. Now we want to find all possible orders of matrix having 5 elements, thus we find all ordered pairs, the product of whose components is 5.
Thus, such possible ordered pairs are (1, 5), (5, 1).
Hence possible orders are 1 x 5 ; 5 x 1.

(ii) We know that, the product of two matrices is possible.
When no. of columns in first matrix = no. of rows in second matrix.
Thus, the product of m x p and p x n matrix is a m x n matrix.

Question 32.
If \(\left[\begin{array}{cc}
x y & 4 \\
z+6 & x+y
\end{array}\right]=\left[\begin{array}{cc}
8 & w \\
0 & 6
\end{array}\right]\), write the value of (x + y + z) = ………………
Solution:
Given
Thus, their corresponding elements are equal.
∴ xy = 8 …(1);
w = 4; z + 6 = 0 ⇒ z = – 6
and x + y = 6 …(2)
from (1) ; y = \(\frac { 8 }{ x }\)
∴ from (2); x + \(\frac { 8 }{ x }\) = 6
⇒ x² – 6x + 8 = 0
⇒ (x – 2) (x – 4) = 0 ⇒ x = 2, 4
When x = 2 ∴y = \(\frac { 8 }{ 2 }\) = 4
When x = 4 ∴ y = 2
When x = 4 ; y = 4 ; z = – 6
Then x + y + z = 1 + 2 – 6 = 0
When x = 2, y = 4 ; z = – 6
Then x + y + z = 2 + 4 – 6 = 0

Question 33.
If 2\(\left[\begin{array}{ll}
1 & 0 \\
0 & x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
1 & 8
\end{array}\right]\), then x + y = ………………..
Solution:
Given, 2\(\left[\begin{array}{ll}
1 & 0 \\
0 & x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
1 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2+y & 0 \\
1 & 2 x+2
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
1 & 8
\end{array}\right]\)
∴ 2 + y = 5 ⇒ y = 3
and 2x + 2 = 8 ⇒ x = 3
∴ x + y = 3 + 3 = 6

Question 34.
If \(\left[\begin{array}{cc}
a+b & 2 \\
5 & b
\end{array}\right]=\left[\begin{array}{ll}
6 & 2 \\
5 & 2
\end{array}\right]\), then find a = ……………
Solution:
Given \(\left[\begin{array}{cc}
a+b & 2 \\
5 & b
\end{array}\right]=\left[\begin{array}{ll}
6 & 2 \\
5 & 2
\end{array}\right]\)
Thus, their corresponding elements are equal.
∴ a + b = 6 … (1)
and b = 2
∴ from (1); a = 6 – 2 = 4

Question 35.
If [2x 4]\(\left[\begin{array}{r}
x \\
-8
\end{array}\right]\) value of x is ……………..
Solution:
Given [2x 4]
⇒ [2x² – 32] = 0 ⇒ 2x² – 32 = 0 ⇒ x = ± 4

Question 36.
A 2 x 2 matrix which is both symmetric and skew symmetric is ……………….
Solution:
Given A is symmetric matrix ∴ A’ = A
and A is skew symmetric ∴ A’ = – A
∴ A = – A ⇒ 2A = O
⇒ A = 0 = \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)

Question 37.
A square matrix A = [a_ij]_nxn is said to be
(i) Symmetric if ………………. and
(ii) skew symmetric if ……………….
Solution:
(i) A’ = A
(ii) A’ = – A

Question 38.
If A and B are square matrices of the same order, then
(i) (AB)’ = ……………….
(ii) (A’)’ = ……………….
Solution:
(i) (AB)’ = B’A’
(ii) (A’)’ = A

Question 39.
If A and B are invertible matrices and AB = BA = I, then (AB)^-1 = ……………….
Solution:
Given AB = BA = I
∴ (AB)^-1 = I^-1 = I

Question 40.
The adjoint of a square matrix is the transpose of the matrix of ……………….
Solution:
cofactors

Question 41.
If A is a square matrix, then A is not invertible if ……………….
Solution:
Since A^-1 = \(\frac{1}{|A|}\) adj A
∴ A^-1 exists if |A| ≠ 0
Thus A is not invertible if | A | = 0

Question 42.
If A is a matrix of order 3 x 3, then | kA | is equal to …………..
Solution:
Then | kA | = kⁿ|A|

Question 43.
If A = \(\left[\begin{array}{ll}
5 & 2 \\
7 & 3
\end{array}\right]\), then (adj A) = ………………
Solution:
Given A = \(\left[\begin{array}{ll}
5 & 2 \\
7 & 3
\end{array}\right]\)
The cofactors of R₁ are ; 3 ; – 7
The cofactors of R₂ are ; – 2 ; 5
∴ adj A = \(\left[\begin{array}{rr}
3 & -7 \\
-2 & 5
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
3 & -2 \\
-7 & 5
\end{array}\right]\)

Question 44.
If \(\left[\begin{array}{c}
x-y-z \\
-y+z \\
z
\end{array}\right]=\left[\begin{array}{l}
0 \\
5 \\
3
\end{array}\right]\), then the values of x, y and z are respectively.
(a) 5, 2, 2
(b) 1, – 2, 3
(c) 0, – 3, 3
(d) 11, 8, 3
Solution:
Given \(\left[\begin{array}{c}
x-y-z \\
-y+z \\
z
\end{array}\right]=\left[\begin{array}{l}
0 \\
5 \\
3
\end{array}\right]\)
∴ x – y – z = 0; – y + z = 5 and z = 3
⇒ – y + 3 = 5
⇒ y = – 2
∴ x + 2 – 3 = 0
⇒ x = 1

Question 45.
If A = \(\left[\begin{array}{ll}
x & 1 \\
1 & 0
\end{array}\right]\) and A² is the identity matrix, then x is equal to
(a) – 1
(b) 0
(c) 1
(d) 2
Solution:
Given A = \(\left[\begin{array}{ll}
x & 1 \\
1 & 0
\end{array}\right]\) and A² = I
⇒ \(\left[\begin{array}{ll}
x & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
x & 1 \\
1 & 0
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
x^2+1 & x \\
x & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
∴ x² + 1 = 1
⇒ x² = 0 ⇒ x = 0

Question 46.
If A = \(\left[\begin{array}{rr}
2 & -1 \\
-1 & 2
\end{array}\right]\) and I is the unit matrix of order 2, then A² equals
(a) 4A – 31
(b) 3A – 4I
(c) A – I
(d) A + I
Solution:
A² = A . A
= \(\left[\begin{array}{rr}
2 & -1 \\
-1 & 2
\end{array}\right]\left[\begin{array}{rr}
2 & -1 \\
-1 & 2
\end{array}\right]=\left[\begin{array}{rr}
5 & -4 \\
-4 & 5
\end{array}\right]\)
Now 4A – 3I = 4\(\left[\begin{array}{rr}
2 & -1 \\
-1 & 2
\end{array}\right]-3\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
8 & -4 \\
-4 & 8
\end{array}\right]-\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]=\left[\begin{array}{rr}
5 & -4 \\
-4 & 5
\end{array}\right]\)

Question 47.
If x\(\left[\begin{array}{r}
-3 \\
4
\end{array}\right]+y\left[\begin{array}{l}
4 \\
3
\end{array}\right]=\left[\begin{array}{r}
10 \\
-5
\end{array}\right]\), then
(a) x = -2, y = 1
(b) x = – 9, y = 10
(c) x = 22, y = 1
(d) x = 2, y = – 1
Solution:
x\(\left[\begin{array}{r}
-3 \\
4
\end{array}\right]+y\left[\begin{array}{l}
4 \\
3
\end{array}\right]=\left[\begin{array}{r}
10 \\
-5
\end{array}\right]\)
⇒ \(\left[\begin{array}{r}
-3 x+4 y \\
4 x+3 y
\end{array}\right]=\left[\begin{array}{r}
10 \\
-5
\end{array}\right]\)
∴ – 3x + 4y = 10 …(1)
and 4x + 3y = – 5 …(2)
3 x eqn. (1) – 4 x eqn. (2); we have
– 25x = 50 ⇒ x = – 2
∴ from (1); 6 + 4y = 10 ⇒ y = 1

Question 48.
If A = \(\left[\begin{array}{cc}
3 & x-1 \\
2 x+3 & x+2
\end{array}\right]\) is a symmetric matrix, then the value of x is
(a) 4
(b) 3
(c) – 4
(d) – 3
Solution:
Given A = \(\left[\begin{array}{cc}
3 & x-1 \\
2 x+3 & x+2
\end{array}\right]\) is symmetric matrix.
∴ A’ = A
⇒ \(\left[\begin{array}{cc}
3 & 2 x+3 \\
x-1 & x+2
\end{array}\right]=\left[\begin{array}{cc}
3 & x-1 \\
2 x+3 & x+2
\end{array}\right]\)
∴ 2x + 3 = x – 1 ⇒ x = – 4

Question 49.
If A is a square matrix, then
(a) A + A^T is symmetric
(b) AA^T is skew symmetric
(c) A^T + A is skew symmetric
(d) A^TA is skew symmetric
Solution:
Let B = A + A^T
∴ B^T = (A + A^T)^T = A^T + (A^T)^T
= A^T + A = A + A^T = B
∴ B is symmetric.

Question 50.
For what value of x is the matrix A = \(\left[\begin{array}{rrr}
0 & 1 & -2 \\
-1 & 0 & 3 \\
x & -3 & 0
\end{array}\right]\) is a skew matrix?
(a) – 2
(b) 0
(c) 2
(d) 3
Solution:
Given A = \(\left[\begin{array}{rrr}
0 & 1 & -2 \\
-1 & 0 & 3 \\
x & -3 & 0
\end{array}\right]\)
and A is skew-symmetric matrix
∴ A’ = – A
⇒ \(\left[\begin{array}{rrr}
0 & -1 & x \\
1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right]=-\left[\begin{array}{rrr}
0 & 1 & -2 \\
-1 & 0 & 3 \\
x & -3 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{rrr}
0 & -1 & x \\
1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right]=\left[\begin{array}{rrr}
0 & -1 & 2 \\
1 & 0 & -3 \\
-x & 3 & 0
\end{array}\right]\)
∴ x = 2

Question 51.
If A = \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
a & b & -1
\end{array}\right]\), then A² is equal to
(a) a null matrix
(b) a unit matrix
(c) – A
(d) A
Solution:
A² = A.A = \(\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
a & b & -1
\end{array}\right]\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
a & b & -1
\end{array}\right]\)
= \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = I₃
which is a unit matrix.

Question 52.
If A + I = \(\left[\begin{array}{rr}
3 & -2 \\
4 & 1
\end{array}\right]\), then (A + I) (A – I) is equal to
(a) \(\left[\begin{array}{rr}
-5 & -4 \\
8 & -9
\end{array}\right]\)
(b) \(\left[\begin{array}{ll}
-5 & 4 \\
-8 & 9
\end{array}\right]\)
(c) \(\left[\begin{array}{ll}
5 & 4 \\
8 & 9
\end{array}\right]\)
(d) \(\left[\begin{array}{ll}
-5 & -4 \\
-8 & -9
\end{array}\right]\)
Solution:
A + I = \(\left[\begin{array}{rr}
3 & -2 \\
4 & 1
\end{array}\right]\)
A – I = A + I – 2 – I
= \(\left[\begin{array}{rr}
3 & -2 \\
4 & 1
\end{array}\right]-2\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & -2 \\
4 & -1
\end{array}\right]\)
∴ (A + I) (A – I) = \(\left[\begin{array}{rr}
3 & -2 \\
4 & 1
\end{array}\right]\left[\begin{array}{rr}
1 & -2 \\
4 & -1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
-5 & -4 \\
8 & -9
\end{array}\right]\)

Question 53.
If A = \(\left[\begin{array}{rr}
1 & -2 \\
4 & 5
\end{array}\right]\) and f(t) = t² – 3t + 7, f(A) + \(\left[\begin{array}{rr}
3 & 6 \\
-12 & -9
\end{array}\right]\) is equal to
(a) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
(b) \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
(c) \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\)
(d) \(\left[\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right]\)
Solution:

Question 54.
Let A = \(\left[\begin{array}{ll}
5 & 0 \\
1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}\right]\). If 4A + 5B – C = 0, then C is
(a) \(\left[\begin{array}{rr}
5 & 25 \\
-1 & 0
\end{array}\right]\)
(b) \(\left[\begin{array}{rr}
20 & 5 \\
-1 & 0
\end{array}\right]\)
(c) \(\left[\begin{array}{cc}
5 & -1 \\
0 & 25
\end{array}\right]\)
(d) \(\left[\begin{array}{rr}
5 & 25 \\
-1 & 5
\end{array}\right]\)
Solution:
Given A = \(\left[\begin{array}{ll}
5 & 0 \\
1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}\right]\)
and 4A + 5B – C = 0
⇒ C = 4A + 5B
⇒ C = 4\(\left[\begin{array}{ll}
5 & 0 \\
1 & 0
\end{array}\right]+5\left[\begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}\right]\)
= \(\left[\begin{array}{cc}
20 & 5 \\
-1 & 0
\end{array}\right]\)

Question 55.
If \(\left[\begin{array}{ll}
x+1 & x-1 \\
x-3 & x+2
\end{array}\right]=\left[\begin{array}{ll}
4 & 1 \\
1 & 3
\end{array}\right]\), then write the value of x.
(a) 4
(b) – 3
(c) 2
(d) – 1
Solution:
Given \(\left[\begin{array}{ll}
x+1 & x-1 \\
x-3 & x+2
\end{array}\right]=\left[\begin{array}{ll}
4 & 1 \\
1 & 3
\end{array}\right]\)
Thus their corresponding entries are equal.
x + 2 = 4 ⇒ x = 2; x – 1 = 1 ⇒ x = 2
x – 3 = – 1 ⇒ x = 2 ; x + 1 = 3 ⇒ x = 2

Question 56.
If A = \(\left[\begin{array}{rr}
1 & 2 \\
3 & -1
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
1 & 3 \\
-1 & 1
\end{array}\right]\) write the value of |AB|.
|AB| = |A| |B| = \(\left|\begin{array}{rr}
1 & 2 \\
3 & -1
\end{array}\right|\left|\begin{array}{rr}
1 & 3 \\
-1 & 1
\end{array}\right|\)
= (- 1 – 6) (1 + 3) = – 7 x 4 = – 28

Question 57.
If for any 2 x 2 square matrix A, A (adj A) = \(\left|\begin{array}{ll}
8 & 0 \\
0 & 8
\end{array}\right|\), then write the value of | A |.
(a) 8
(b) 16
(c) 0
(d) 64
Solution:
Given A adj A = \(\left[\begin{array}{ll}
8 & 0 \\
0 & 8
\end{array}\right]\)
= 8 \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) = 8 I₂
We know that,
A adjA = |A | I₂
∴ | A | = 8

Question 58.
If A is a square non-singular matrix of order 3 such that | adj A | = 64, find | A |.
(a) 32
(b) 4
(c) ±8
(d) None of these
Solution:
We know that,
If A be a square matrix of order n.
Then | adj A | = | A |^n-1
Given A be a non-singular matrix of order 3
Then | adj A | = | A |^3-1 = | A |²
also, | adj A | = 64
∴ | A |² = 64
∴ | A | = ± 8

Question 59.
For what value of x, is the given matrix singular ?
(a) 1
(b) – 1
(c) 2
(d) None of these
Solution:
We know that a matrix A is singular
∴ | A | = 0
⇒ \(\left|\begin{array}{cc}
3-2 x & x+1 \\
2 & 4
\end{array}\right|\)
⇒ 4 (3 – 2x) – 2 (x + 1) = 0
⇒ 10 – 10x = 0
⇒ x = 1

Question 60.
Write A^-1 for A = \(\left[\begin{array}{rr}
2 & 3 \\
5 & -2
\end{array}\right]\).
Solution:
(a) \(\left[\begin{array}{rr}
2 & 5 \\
3 & -2
\end{array}\right]\)
(b) \(\left[\begin{array}{rr}
-2 & -3 \\
-5 & 2
\end{array}\right]\)
(c) \(\frac{1}{19}\left[\begin{array}{rr}
-2 & -5 \\
-3 & 2
\end{array}\right]\)
(d) \(\frac{1}{19}\left[\begin{array}{rr}
2 & 3 \\
5 & -2
\end{array}\right]\)
Solution:

Question 61.
Which of the following is correct?
(a) Determinant is a square matrix.
(b) Determinant is a number associated to a matrix.
(c) Determinant is a number associated to a square matrix.
(d) All of the above.
Solution:
(c) Determinant is a number associated to a square matrix.

Question 62.
If A = \(\left[\begin{array}{rr}
\log x & -1 \\
-\log x & 2
\end{array}\right]\) and if det (A) = 2, then the value of x is equal to
(a) 2
(b) e²
(c) – 2
(d) e
(e) log 2
Solution:
Given A = \(\left[\begin{array}{rr}
\log x & -1 \\
-\log x & 2
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{rr}
\log x & -1 \\
-\log x & 2
\end{array}\right|\)
= 2 log x – log x = log x
and det (A) = 2
∴ log x = 2 ⇒ x = e²

Question 63.
If \(\left[\begin{array}{cc}
x+y & x-y \\
2 x+z & x+z
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
1 & 1
\end{array}\right]\) then the value of x, y, z are respectively
(a) 0, 0, 1
(b) 1, 1, 0
(c) 0, 0, 0
(d) 1, 1, 1
Solution:
Given \(\left[\begin{array}{cc}
x+y & x-y \\
2 x+z & x+z
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
1 & 1
\end{array}\right]\)
∴ x + y = 0 …(1)
x – y = 0 …(2)
2x + z = 1 …(3)
x + z = 1 …(4)
On solving eqns. (1) and (2); we have x = y = 0
∴ from (3); z = 1

Question 64.
If [x 1] \(\left[\begin{array}{rr}
1 & 0 \\
-2 & 0
\end{array}\right]\) = 0, then x equals
(a) 0
(b) – 2
(c) – 2
(d) 2
Solution:
[x 1]_1×2 \(\left[\begin{array}{rr}
1 & 0 \\
-2 & 0
\end{array}\right]\)_2×2 = 0
⇒ \(\left[\begin{array}{c}
x-2 \\
0
\end{array}\right]=\left[\begin{array}{l}
0 \\
0
\end{array}\right]\)
∴ x – 2 = 0
⇒ x = 2

Question 65.
\(\left[\begin{array}{lll}
7 & 1 & 5 \\
8 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
2 \\
3 \\
1
\end{array}\right]+5\left[\begin{array}{l}
1 \\
0
\end{array}\right]\) is equal to
(a) \(\left[\begin{array}{l}
16 \\
27
\end{array}\right]\)
(b) \(\left[\begin{array}{l}
27 \\
16
\end{array}\right]\)
(c) \(\left[\begin{array}{c}
15 \\
16
\end{array}\right]\)
(d) \(\left[\begin{array}{l}
16 \\
15
\end{array}\right]\)
(e) \(\left[\begin{array}{l}
16 \\
16
\end{array}\right]\)
Solution:
\(\left[\begin{array}{lll}
7 & 1 & 5 \\
8 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
2 \\
3 \\
1
\end{array}\right]+5\left[\begin{array}{l}
1 \\
0
\end{array}\right]\)
= \(\left[\begin{array}{l}
14+3+5 \\
16+0+0
\end{array}\right]+\left[\begin{array}{l}
5 \\
0
\end{array}\right]=\left[\begin{array}{l}
22 \\
16
\end{array}\right]+\left[\begin{array}{l}
5 \\
0
\end{array}\right]=\left[\begin{array}{l}
27 \\
16
\end{array}\right]\)

Question 66.
If the square of the matrix \(\left[\begin{array}{rr}
a & b \\
a & -a
\end{array}\right]\)
is the unit matrix, then b is equal to
(a) \(\frac{a}{1+a^2}\)
(b) \(\frac{1-a^2}{a}\)
(c) \(\frac{1+a^2}{a}\)
(d) \(\frac{a}{1-a^2}\)
Solution:
Let A = \(\left[\begin{array}{rr}
a & b \\
a & -a
\end{array}\right]\) s.t A² = I
⇒ \(\frac { 1 }{ 2 }\)
⇒ \(\frac { 1 }{ 2 }\)
∴ a² + ab = 1 ⇒ b = \(\frac { 1 }{ 2 }\)

Question 67.
(i) If A = \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\), then A² is equal to
(a) \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\)
(b) \(\left[\begin{array}{ll}
1 & 0 \\
1 & 0
\end{array}\right]\)
(c) \(\left[\begin{array}{ll}
0 & 1 \\
0 & 1
\end{array}\right]\)
(d) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
(ii) Alternative Form. If matrix A = [a_ij]_2×2, where a_ij= 1, if i ≠ j and = 0 if i = j, then A² is equal to (a) I
(b) A
(c) O
(d) None of these
Solution:
(i) A² = A . A
= \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) = I

(ii) A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]=\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\)
∴ A² = \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= I

Question 68.
If U = [2 – 1 4], X = [0 2 3], V = \(\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\) and X = \(\left[\begin{array}{l}
2 \\
2 \\
4
\end{array}\right]\), then UV + XY is equal to
(a) [24]
(b) 20
(c) [- 20]
(d) – 20
Solution:
UV + XY = \(\left[\begin{array}{lll}
2 & -1 & 4
\end{array}\right]\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]+\left[\begin{array}{lll}
0 & 2 & 3
\end{array}\right]\)
\(\left[\begin{array}{l}
2 \\
2 \\
4
\end{array}\right]\)
= [6 – 2 + 4] + [0 + 4 + 12] = [8] + [16]
= [24]

Question 69.
If P = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
1 & 3 & 1
\end{array}\right]\), Q = PP^T, then the value of the determinant of Q is equal to
(a) 2
(b) – 2
(c) 1
(d) 0
Solution:

Question 70.
If A = \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right]\),
then det (adj A) equals
(a) a²⁷
(b) a⁹
(c) a⁶
(d) a²
Solution:
A = \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right|\)
Expanding along R₁
= a\(\left|\begin{array}{ll}
a & 0 \\
0 & a
\end{array}\right|\) = a x a² = a³
If A be a matrix of order n.
Then | adj A | = | A |^n-1
Here A be a matrix of order 3.
Then | adj A | = | A |^3-1 = (a³)² = a⁶

Question 71.
If A is a 3 x 3 matrix such that | A | = 8, then | 3A | is equal to
(a) 8
(b) 24
(c) 72
(d) 216
Solution:
(d) 216
We know that if A be a square matrix of order n. Then det (kA) = kⁿ det A
∴ | 3A | = 3³ | A | = 27 x 8 = 216

Question 72.
If A is a square matrix of order 3 such that A (adj A) = 10 I, then | adj A | is equal to
(a) 1
(b) 10
(c) 100
(d) 101
Solution:
Given A be a square matrix of order 3
s.t A (adj A) = 101
also, A (adj A) = | A |I
∴ | A | = 10
Thus | adj A | = | A |^3-1 = | A |² = 10² = 100

Question 73.
If A = \(\left[\begin{array}{cc}
2-k & 2 \\
1 & 3-k
\end{array}\right]\) is a singular matrix, then the value of 5k – k³ is
(a) 0
(b) 6
(c) – 6
(d) 4
Solution:
Given A = \(\left[\begin{array}{cc}
2-k & 2 \\
1 & 3-k
\end{array}\right]\) is a singular matrix
∴ | A | = 0 ⇒ \(\left|\begin{array}{cc}
2-k & 2 \\
1 & 3-k
\end{array}\right|\) = 0
⇒ (2 – k)(3 – k) – 2 = 0 ⇒ k² – 5k + 4 = 0
⇒ (k – 1) (k – 4) = 0 ⇒ k = 1, 4
When k = 1 ; ∴ 5k – k³ = 5 x 1 – 1³ = 4
When k = 4 ; ∴ 5k – k³ = 20 – 64 = – 44

Question 74.
If A^-1 = \(\left[\begin{array}{rr}
5 & -2 \\
-7 & 3
\end{array}\right]\) and B^-1 = \(\frac{1}{2}\left[\begin{array}{rr}
9 & -7 \\
-8 & 6
\end{array}\right]\) then (AB)^-1 = ?
(a) \(\left[\begin{array}{cc}
94 & -39 \\
-82 & 34
\end{array}\right]\)
(b) \(\left[\begin{array}{rr}
94 & -82 \\
-39 & 34
\end{array}\right]\)
(c) \(\left[\begin{array}{rr}
-47 & 46 \\
-\frac{39}{2} & -17
\end{array}\right]\)
(d) \(\left[\begin{array}{rr}
47 & -\frac{39}{2} \\
-41 & 17
\end{array}\right]\)
Solution:
Given A^-1 = \(\left[\begin{array}{rr}
5 & -2 \\
-7 & 3
\end{array}\right]\) and B^-1 = \(\frac{1}{2}\left[\begin{array}{rr}
9 & -7 \\
-8 & 6
\end{array}\right]\)
∴ (AB)^-1 = B^-1A^-1 = \(\frac{1}{2}\left[\begin{array}{rr}
9 & -7 \\
-8 & 6
\end{array}\right]\left[\begin{array}{rr}
5 & -2 \\
-7 & 3
\end{array}\right]=\frac{1}{2}\left[\begin{array}{rr}
94 & -39 \\
-82 & 34
\end{array}\right]=\left[\begin{array}{rr}
47 & -\frac{39}{2} \\
-41 & 17
\end{array}\right]\)

Question 75.
If A = \(\left[\begin{array}{ll}
2 & 1 \\
0 & x
\end{array}\right]\) and A^-1 = \(\left[\begin{array}{cc}
1 / 2 & 1 / 6 \\
0 & 1 / x
\end{array}\right]\), then the value of x is equal to
(a) – 3
(b) 3
(c) – 2
(d) 6
(e) – 6
Solution:
We know that AA^-1 = I
⇒ \(\left[\begin{array}{cc}
2 & 1 \\
0 & x
\end{array}\right]\left[\begin{array}{cc}
1 / 2 & 1 / 6 \\
0 & 1 / x
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \quad \Rightarrow\left[\begin{array}{cc}
1 & \frac{1}{3}+\frac{1}{x} \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
∴ \(\frac { 1 }{ 3 }\) + \(\frac { 1 }{ x }\) = 0 ⇒ x = – 3

Question 76.
If A = \(\left[\begin{array}{rr}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), then AA^-1 =
(a) A
(b) zero matrix
(c) A’
(d) I
Solution:
AA’ = \(\left[\begin{array}{rr}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\cos ^2 \alpha+\sin ^2 \alpha & -\cos \alpha \sin \alpha+\sin \alpha \cos \alpha \\
-\sin \alpha \cos \alpha+\cos \alpha \sin \alpha & \sin ^2 \alpha+\cos ^2 \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) = I

Question 77.
Matrix A = \(\left[\begin{array}{ccr}
0 & 2 b & -2 \\
3 & 1 & 3 \\
3 a & 3 & -1
\end{array}\right]\) is given to be symmetric, then the values of a and b respectively are
(a) \(\frac{3}{2},-\frac{2}{3}\)
(b) \(-\frac{2}{3}, \frac{3}{2}\)
(c) 2, – 3
(d) 3, – 2
Solution:
Given A = \(\left[\begin{array}{ccr}
0 & 2 b & -2 \\
3 & 1 & 3 \\
3 a & 3 & -1
\end{array}\right]\) is given to be symmetric
∴ A’ = A
⇒ \(\left[\begin{array}{rrr}
0 & 3 & 3 a \\
2 b & 1 & 3 \\
-2 & 3 & -1
\end{array}\right]=\left[\begin{array}{ccr}
0 & 2 b & -2 \\
3 & 1 & 3 \\
3 a & 3 & -1
\end{array}\right]\)
Thus their corresponding entries are equal.
∴ 3 = 2b ⇒ b = \(\frac { 3 }{ 2 }\)
and 3a = – 2 ⇒ a = – \(\frac { 2 }{ 3 }\)

Question 78.
In the matrix A = \(\left[\begin{array}{ccc}
a & 1 & x \\
2 & \sqrt{3} & x^2-y \\
0 & 5 & -\frac{2}{5}
\end{array}\right]\),
write (i) the order of the matrix A (ii) the number of the element (iii) elements a₂₃, a₃₁ and a₁₂
Solution:
Given A = \(\left[\begin{array}{ccc}
a & 1 & x \\
2 & \sqrt{3} & x^2-y \\
0 & 5 & -\frac{2}{5}
\end{array}\right]\)
(i) matrix A has 3 rows and 3 columns and it is of order 3×3
(ii) matrix A have 3 x 3 i.e. 9 elements.
(iii) a₂₃ = x² – y ; a₃₁ = 0 ; a₁₂ = 1

Question 79.
(i) If a matrix has 5 elements, write all possible orders it can have.
(ii) If A is matrix of order 3×4 and B is a matrix of order 4×3, then the order of matrix (AB) is ………….
Solution:
(i) We know that, if a matrix of order m x n then it has mn elements. Thus to find possible orders of matrix containing 5 elements we have to find all ordered pairs (a, b) such that a and b are factors of 5. Hence such possible ordered pairs are (1, 5) and (5, 1).
∴ possible orders are 1 x 5 and 5×1.

(ii) Given A is a matrix of order 3×4 and B be a matrix of 4 x 3.
Thus, the order of matrix AB is 3 x 3.

Question 80.
If matrix A = [1 2 3], then find AA’ where A’ is the transpose of matrix A.
Solution:
Given A = [1 2 3] and A’ = \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
∴ AA’ = \(\left[\begin{array}{lll}
1 & 2 & 3
\end{array}\right]_{1 \times 3}\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]_{3 \times 1}\)
= [1 + 4 +9] = [14]

Question 81.
(i) Write the number of all possible matrices of order 2×2 with each entry 1, 2 or 3.
(ii) Similar Question. The number of 3 x 3 matrices with entries – 1 or + 1 is
(a) 2⁴
(b) 2⁵
(c) 2⁶
(d) 2⁹
Solution:
(i) Let A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]_{2 \times 2}\)
since each entry is 1, 2 or 3. i.e. each entry is filled with 1, 2 or 3.
Thus, each a_ij can be filled by 3 ways and there are 4 elements.
Hence the total number of 2 x 2 matrices with each entry is either 1, 2 or 3 = 34 = 81.

(ii) Now the matrices of order 3×3 contains 9 entries and each entry can be 1 or – 1 i.e. there are two ways to fill each entry of given matrix.
∴ Total no. of such matrices = 29

Question 82.
If A is a skew matrix of order 3, then prove that det A = 0. [Remember this result]
Solution:
Given A be a skew symmetric matrix of order 3 x 3
∴ A’ = – A ⇒ | A’| = |- A | = (- 1)³ | A | [If A be a matrix of order n x n Then | AA | = kⁿ | A |]
⇒ | A | + | A | = 0 [∵ | A’ | = | A | ]
⇒ 2 | A | = 0 ⇒ | A | = 0

Question 83.
If A is a square matrix such that A² = I, then find the simplified value of (A – I)³ + (A + I)³ – 7 A.
Solution:
(A – I)³ = (A-I)(A-I)² = (A-I)(A-I) (A-I) = (A-I) (A²- AI – IA + I²)
= (A – I)(I – A – A + I)
[∵ A² = I and IA = A]
= (A – I) (2I – 2A) =- 2 (A²-AI-IA + I²) = – 2 (I – A – A + I)
= – 2 (2I – 2A) = + 4 (A – I)
(A + I)³ = (A + I) (A + I) (A + I)
= (A + I) (A² + AI + IA + I²)
= (A + I) (I + A + A + I)
= 4 (A + I)
∴ (A – I)³ + (A + I)³ – 7A = 4 (A – I) + 4 (A + I) – 7A = A

Question 84.
Verify that A² = I, when
A = \(\left[\begin{array}{rrr}
0 & 1 & -1 \\
4 & -3 & 4 \\
3 & -3 & 4
\end{array}\right]\)
Solution:
A² = \(\left[\begin{array}{rrr}
0 & 1 & -1 \\
4 & -3 & 4 \\
3 & -3 & 4
\end{array}\right]\)
= \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = I

Question 85.
If A and B are matrices of order 3 and | A | = 5, | B | = 3, then find the value of | 3AB |.
Solution:
Given A and B be matrices of order 3 then AB be also a matrix of order 3.
Then | A | = 5 ; | B | = 3
∴ | 3AB | = 3³ | A | | B | = 27 x 5 x 3 = 405
[if A be a matrix of order n. Then | kA | = kⁿ|A| ]

Question 86.
If A is a square matrix of order 3, with | A | = 9, then write the value of | 2 adj A|.
Solution:
If A be a matrix of order 3. Then adj A is also a matrix of order 3.
Then | 2 adj A | = 2³ | adj A |
= 8 | A |^3-1 = 8 x | A |²
= 8 x 9² = 8 x 81 = 648

Question 87.
If A and B are square matrices of the same order 3, such that | A | = 2 and AB = 2 1, write the value of | B |.
Solution:
Given | A | = 2 and AB = 2I
⇒ | AB | = | 2I | ⇒ | A||B | = 2³ |I| ⇒ 2 x |B| = 8 x 1 ⇒ | B | = 4

Question 88.
If A is a non-singular matrix of order 3 and | adj A | = | A |^k, then write the value of k.
Solution:
If A be a non-singular matrix of order n.
Then | Adj A | = | A |^n-1
also given | adj A | = | A |^k
Here n = 3 ; | adj A | = | A |^3-1 = | A |²
∴ k = 2

Question 89.
(i) If A is an invertible matrix of order 2 and det (A)=4, then write the value of det (A^-1 ).
(ii) If A is a 3 x 3 invertible matrix then, what will be the k if det (A^-1) = (det A)^k .
Solution:
(i) Given A be an invertible matrix of order 2
∴ | A | ≠ 0 and | A | = 4
det (A^-1) = | A^-1 | = | A|^-1 = 4^-1 = \(\frac{1}{4}\)

(ii) If A be an invertible matrix of order 3 s.t | A | ≠ 0
Given det (A^-1) = (det A)^k
Since, det (A^-1) = (det A)^k
∴ k = – 1

Question 90.
Given A = \(\left[\begin{array}{rrr}
4 & 2 & 5 \\
2 & 0 & 3 \\
-1 & 1 & 0
\end{array}\right]\), value of det. (2 AA^-1)
Solution:
Here |A| = \(\left|\begin{array}{rrr}
4 & 2 & 5 \\
2 & 0 & 3 \\
-1 & 1 & 0
\end{array}\right|\);
Expanding along R₁ = 4 (0 – 3) – 2 (0 + 3) + 5(2 – 0) = – 12 – 6 + 10 = – 8 ≠ 0
∴ A^-1 exists
We know that AA^-1 = I = A^-1A
∴ det (2 A A^-1) = det (2I)
= 2³ det (I) = 8 x 1=8
[if A be a matrix of order n. Then | kA | = kⁿ|A|]

Practicing S Chand Class 12 Maths Solutions Chapter 25 Application of Integrals Ex 25(a) is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 12 ICSE Maths Solutions Chapter 25 Application of Integrals Ex 25(a)

Question 1.
Using integration, find the area of the region bounded by
(i) the line 3 y = 2 x + 4, the x-axis and the lines x = 1 and x = 3.
(ii) 2 y = -x + 6, the x-axis and the lines x = 2 and x = 4.
Answer:
(i) The eqn. of given line be
3 y = 2 x + 4 ……………..(1)
eqn. (1) meets x-axis at A(-2, 0) and y-axis at B (0, \(\frac{4}{3}\)).
∴ required area = \(\int_1^3\) y d x
= \(\int_1^3\) \(\frac{2 x+4}{3}\) d x = \(\frac{1}{3}\) \(\frac{(2 x+4)^2}{2 \times 2}]_1^3\)
= \(\frac{1}{12}\) [(10)² – 6²]
= \(\frac{1}{12}\)(100 – 36)
= \(\frac{64}{12}\) = \(\frac{16}{3}\) sq. units

(ii) The eqn. of given line be
2y = -x + 6
it meets coordinate axes at A(6, 0) and B(0, 3).
∴ Required area = \(\int_2^4\) y d x
= \(\int_2^4\) \(\frac{-x+6}{2}\) d x
= \(\frac{1}{2}\) \(\frac{(6-x)^2}{-2}]_2^4 \)
= \( -\frac{1}{4}\) [4 – 16]
= 3 sq. units

Question 2.
(i) Using integration, find the area of the region bounded between the line x = 2 and the parabola y² = 8 x.
(ii) Find the area of the region bounded by the parabola y² = 12 x and its latus rectum.
(iii) Find the area enclosed between the co-ordinate axes and the curve y² = 4 a(x + λ) in the second quadrant.
Answer:
(i) Clearly the figure is symmetrical about x-axis.
∴ area enclosed by the parabola
= 2 × area enclosed by parabola
and x = 2 in Ist quadrant
Divide the region in first quadrant into vertical strips each vertical strip has its lower end on x-axis and upper and on the parabola y = \(\sqrt{8 x}\). So the approximating rectangle has length |y| and width d x. This rectangle can move between x = 0 and x = 2.

[∵ y ≥ 0 ∴|y|=y]
∴ required area = 2 × \(\int_0^2\)|y| d y
= 2 \(\int_0^2\) y d y
= 2 \(\int_0^2\) √ 8 √ x d x
= 2 × 2 \(\sqrt{2}\) [\(\frac{2}{3}\) x^3/2 ]² ₀
= \(\frac{8 \sqrt{2}}{3}\) [2^3/2 – 0]
= \(\frac{32}{3}\) sq. units

(ii) Given eqn. of parabola be y² = 12 x represents a right handed parabola with vertex at (0,0). Here 4 a = 12
⇒ a = 3
∴ required area = 2 × area of region OABO
= 2 × \(\int_0^3\) \(\sqrt{12 x}\) d x
= 2 × 2 \(\sqrt{3}\) \(\int_0^3\) x^1/2 d x
= 4 \(\sqrt{3}\) \(\frac{x^{3 / 2}}{3 / 2}]_0^3\)
= \(\frac{8 \sqrt{3}}{3}\)[3 \(\sqrt{3}\) – 0]
= 24 sq. units

(iii) The given eqn. of parabola be
y² = 4 a(x + λ)
represents a right handed parabola with vertex (-λ, 0) it meets y-axis at A(0, 2 \(\sqrt{a \lambda}\))
∴ required area = \(\int_{-\lambda}^0 2 \sqrt{a}\) \(\sqrt{x+\lambda}\) d x
[∵ y > 0 in 2nd quad ]
= 2 \(\sqrt{a}\) \(\frac{(x+\lambda)^{3 / 2}}{3 / 2}]_{-\lambda}^0\)
= \(\frac{4 \sqrt{a}}{3}\) λ^3/2

Question 3.
Draw the rough sketch of the curve y = \(\sqrt{3 x + 4 x}\) and find the area under the curve, above the x-axis and between x = 0 and x = 4.
Answer:
The eqn. of given curve y = \(\sqrt{3 x+4}\)
The table of values is given as under :

Plot the points (0, 2) ;(4, 4) and (2, \(\sqrt{10}\)) on graph paper and join then by free hand curve.
∴ Required area of region OABCO
= \(\int_0^4\) \(\sqrt{3 x+4}\) d x
= \(\frac{(3 x+4)^{3 / 2}}{\frac{3}{2} \cdot 3}]_0^4\)
= \(\frac{2}{9}[(16)^{3 / 2}-(4)^{3 / 2}]\)
= \(\frac{2}{9}\)[64 – 8]
= \(\frac{112}{9}\) sq. units

Question 4.
Sketch the graph of y = |x + 1|. Evaluate \(\int_{-4}^2\)|x + 1| d x. What does the value of the integral represent on the graph?
Answer:
For Graph ; For x < -1; y = f(x) = -(x + 1) For x = -1 ; y = 0 For x > -1 ; y = x + 1
∴ \(\int_{-4}^2\)|x + 1| d x
= \(\int_{-4}^{-1}\) – (x + 1) d x + \(\int_{-1}^2\) (x + 1) d x
∴ I = \(-\frac{(x+1)^2}{2}]_{-4}^1\) + \(\frac{(x+1)^2}{2}]_{-1}^2\)
= \(-\frac{1}{2}\)(0 – 9) + \(\frac{1}{2}\)(9 – 0)
= \(\frac{9}{2}\) + \(\frac{9}{2}\) = 9

When x = -4 < -1 ∴ y = -(-4 + 1) = 3 i.e. point becomes (-4, 3). When x = 2 > -1
∴ y = 2 + 1 = 3 i.e. point becomes (2, 3).
Further, area of ∆ ABC = \(|\frac{1}{2} \times(-3) \times(3)|\) = \(\frac{9}{2}\) sq. units
and area of ∆CDE = \(|\frac{1}{2} \times 3 \times 3|\) = \(\frac{9}{2}\) sq. units
∴ \(\int_{-4}^2\)|x + 1| d x
= sum of areas of two ∆ from -4 to -1 and -1 to 2 .

Question 5.
Find the area bounded by the curve y = x² and the line y = 16. [Fig. 25.41]

Answer:
(i) The given curve y = x²
and the line y = 16 intersects when 16 = x²
⇒ x = ± 4
∴ Region R = {(x, y) ; x² ≤ y ≤ 10 ; 0 ≤ x ≤ 4}
∴ Required area = \(\int_0^4\)[16 – x² ] d x
= 16 x – \(\frac{x^3}{3}]_0^4\)
= 64 – \(\frac{64}{3}\)
= \(\frac{128}{3}\) sq. units

(ii) required area = \(\int_1^4\) x d y
= \(\int_1^4\) \(\frac{\sqrt{y}}{2}\) d y
= \(\frac{1}{2}\) \(\frac{y^{3 / 2}}{3 / 2}]_1^4\)
= \(\frac{1}{3}\)[4^3/2 – 1] = \(\frac{7}{3}\) sq. units

Question 6.
Sketch the region lying in the first quadrant and bounded by y = 4 x², x = 0, y = 1 and y = 4. Find the area of the region, using integration.

Answer:
The given curve is y = 4 x² be an upward parabola with vertex at (0, 0) and is symmetrical about y-axis.
∴ required area = area of region OABCO
= \(\int_1^4\) x d y [Taking Horizontal strips]

= \(\int_1^4 \) \(\frac{\sqrt{y}}{2}\) d y
[∵ y = 4 x² ⇒ x = ± \(\frac{\sqrt{y}}{2}\) as region lies in 1st quadrant ∴ x > 0 i.e. x = \(\frac{\sqrt{y}}{2}\)]
= \(=\frac{1}{2}\) \(\frac{y^{3 / 2}}{3 / 2}]_1^4\)
= \(\frac{1}{3}\)[8 – 1]
= \(\frac{7}{3}\) square units

Question 7.
Find the gradients of the curve y = x²(2 – x) at the points (0, 0) and (2, 0) and sketch the part of the curve of which both x and y are positive. Find the area between this part of the curve and the axis.
Answer:
Given eqn. of curve y = x²(2 – x)
∴ \(\frac{d y}{d x}\) = 4 x – 3 x²
Thus gradient of curve at (0, 0) = \(\frac{d y}{d x}\)_(0,0) = 0
and gradient of curve at (2, 0) = \(\frac{d y}{d x}\)_(2,0)
= 8 – 12 = -4

The table of values is given as under:

by plotting all these points on graph paper and join them by free hand curve.
∴ required area of region OBAO = \(\int_0^2\) x²(2 – x) d x
= \(\frac{2 x^3}{3}\) – \(\frac{x^4}{4}]_0^2\)
= \(\frac{16}{3}\) – 4
= \(\frac{4}{3}\) sq. units

Question 8.
Find the area included between the curve y = 9 – x² , the X-axis and the lines x = -2 and x = ± 2.
Answer:
The given curve
y = 9 – x²
x² = 9 – y = -(y – 9)
⇒ represents a downward parabola with vertex (0, 9).
∴ required area of region P A Q P
= \(\int_{-2}^2\) y d x
= \(\int_{-2}^2\)(9 – x²) d x
= 9 x – \(\frac{x^3}{3}]_{-2}^2\)
= (18 – \(\frac{8}{3}\)) – (-18 + \(\frac{8}{3}\))
= 36 – \(\frac{16}{3}\)
= \(\frac{108-16}{3}\)
= \(\frac{92}{3}\) sq. units

Question 9.
Find the area bounded by the curve y² = 4 x, the line y = 3, and the y-axis.
Answer:
The given curve y² = 4 x represents a right handed parabola with vertex at origin O(0, 0). Now the line y = 3 meets the given curve y² = 4 x when 9 = 4 x
i.e. x = \(\frac{9}{4}\) i.e. at point (\(\frac{9}{4}\), 3).
∴ R = {(x, y) ; 0 ≤ x ≤ \(\frac{y^2}{4}\) ; 0 ≤ y ≤3}
Thus required area = \(\int_0^3\) \(\frac{y^2}{4}\) d y
= \(\frac{1}{4}\) \(\frac{y^3}{3}]_0^3\)
= \(\frac{1}{12}\) × 27
= \(\frac{9}{4}\) sq. units

Aliter: Divide the region into vertical strips with upper end on line y = 3
and. low erend on given curve y² = 4 × and 0 ≤ x ≤ \(\frac{9}{4}\)
∴ required area = \(\int_0^{9 / 4}\)[3 – 2 \(\sqrt{x}\)] dx = 3x – \(\frac{2 x^{3 / 2}}{3 / 2}]_0^{9 / 4}\)
= 3(\(\frac{9}{4}\) – 0) – \(\frac{4}{3}\)[\(\frac{9}{4})^{3 / 2}\) – 0]
= \(\frac{27}{4}\) – \(\frac{4}{3}\) × \(\frac{27}{8}\)
= \(\frac{27}{4}\) – \(\frac{9}{2}\)
= \(\frac{27-18}{4}\) = \(\frac{9}{4}\) sq. units

Question 10.
Find the area of the region bounded by y = -1, y = 2, x = y and x = 0.
Answer:
Given lines are
y = -1 ……………….(1)
y = 2 ……………….(2)
x = y ……………….(3)
x = 0 …………………(4)
and
Since y = -1 and y = 2 are the lines are parallel to x-axis and x = 0 represents y-axis.
The line y = x passes through origin and making a slope of 45° with positive direction of x-axis.
Divide the region into horizontal strips as shown in shaded portion.

R₁ = {(x, y); 0 ≤ x ≤ y ; 0 ≤ y ≤ 2}
R₂ = {(x, y) ; y ≤ x ≤ 0 ;-1 ≤ y ≤ 0}
∴ Required area = \(\int_{-1}^0\)|x| d y + \(\int_0^2\) x d y
= \(\int_{-1}^0\) – y d y + \(\int_0^2\) y d y
= \(-\frac{y^2}{2}]_{-1}^0\) + \(\frac{y^2}{2}]_0^2\)
= \(-\frac{1}{2}\)(0 – 1) + \(\frac{1}{2}\) (4 – 0)
= \(\frac{1}{2}\) + 2 = \(\frac{5}{2}\) sq. units

Question 11.
Find the area bounded by the parabola y² = 2 x and the ordinates x = 1 and x = 4.
Answer:
The given parabola y² = 2 x be a right handed parabola with vertex (0, 0).
Divide the region into vertical strips with lower end on x-axis and upper end on given curve
y² = 2 x .
∴ required area = 2 \(\int_1^4\) y d x
= 2 \(\int_1^4\) \(\sqrt{2}\) \(\sqrt{x}\) d x
= 2 \(\sqrt{2}\) \(\frac{x^{3 / 2}}{3 / 2}]_1^4\)
= \(\frac{4 \sqrt{2}}{3}\)[4^3/2 – 1]
= \(\frac{4 \sqrt{2}}{3}\)[8 – 1]
= \(\frac{28 \sqrt{2}}{3}\) sq.units

Question 12.
Find the area bounded by the curve y² = 4 a² (x – 1) and the lines x = 1 and y = 4 a.
Answer:
Given eqn. of curve be y² = 4 a² (x – 1) which represents a right handed parabola with vertex (1, 0).
The line x = 1 meets given parabola at A(1, 0).
The line y = 4 a meets given parabola when 16 a² = 4 a² (x – 1)
⇒ x = 5 i.e. at B(5, 4 a).
∴ R = {(x, y) ; 4 a ≤ y ≤ 2 a \(\sqrt{x-1}\) ; 1 ≤ x ≤ 5}
Thus required area = \(\int_1^5\) [4 a – 2 a \(\sqrt{x-1}\)] d x
= 4 a x – \(\frac{2 a(x-1)^{3 / 2}}{3 / 2}]_1^5\)
= 4 a(5 – 1) – \(\frac{4 a}{3}\) [4^3/2 – 0]
= 16 a – \(\frac{32 a}{3}\)
= \(\frac{16 a}{3}\) sq.units

Question 13.
Evaluate the area of the region bounded by the curve y = 2 \(\sqrt{1-x^2}\), and the X-axis, after drawings rough sketch of the same.
Answer:
Given equation of curve be y = 2 \(\sqrt{1-x^2}\)
⇒ y² = 4 (1 – x²)
⇒ \(\frac{y^2}{4}\) + \(\frac{x^2}{1}\) = 1
So it represents an ellipse with major axis on y-axis and minor axis as x-axis. Clearly it contains even power of x and y so the given curve is symmetrical about both coordinate axis.

∴ area of whole ellipse = 4 × area of ellipse in first quadrant and coordinate axes.
Now, we divide the region in first quadrant into vertical strips each vertical strip has lower and on x-axis and upper end on ellipse. So the approximating rectangle has length |y| and width d x and clearly it move from x = 0 and x = 1.
∴ area of whole ellipse = 4 × \(\int_0^1\) |y| d x
= 4 \(\int_0^1\) y d x [∵ y ≥ 0]
= 4 \(\int_0^1\) 2 \(\sqrt{1-x^2}\) d x
= 8 \(\frac{x \sqrt{1-x^2}}{2}\) + \(\frac{1}{2}\) sin^-1 x]¹ ₀
= 8[0 + \(\frac{1}{2}\) sin^-1 (1) – 0 – 0]
= 8 × \(\frac{1}{2}\) × \(\frac{\pi}{2}\)
= 2 π square units
Here required area = area of region in first quadrant = \(\frac{1}{4}\) × 2π = \(\frac{\pi}{2}\) square units

Question 14.
Make a rough sketch of the curve 3 y = (2 x + 1)(2 – x), and calculate its gradient at the point where it meets the axis of y. Find the area bounded by the curve and the axis of x.
Answer:
Given eqn. of curve be
3y = (2 x + 1)(2 – x) ……………….(1)
The table of values is given as under :

The given eqn. (1) can be written as,
3y = -2 x² + 3 x + 2
⇒ 3y = -2[x² – \(\frac{3}{2}\) x + \(\frac{9}{16}\) – \(\frac{9}{16}\) – 1]
⇒ 3y = -2[(x – \(\frac{3}{4}\))² – \(\frac{25}{16}\)]
⇒ (x – \(\frac{3}{4}\))²
= \(\frac{25}{16}\) – \(\frac{3}{2}\)
y = \(-\frac{3}{2}\)(y – \(\frac{25}{16}\) × \(\frac{2}{3}\))
⇒ (x – \(\frac{3}{4}\))²
= \(-\frac{3}{2}\) (y – \(\frac{25}{24}\))
which represents a downward parabola with vertex (\(\frac{3}{4}\), \(\frac{25}{24}\)).
Diff. (1) both sides w.r.t. x, we have
3 \(\frac{d y}{d x}\) = -4 x + 3
⇒ \(\frac{d y}{d x}\) = \(\frac{-4 x+3}{3}\)
curve given by eqn. (1) meets y-axis i.e. x = 0
∴ y = \(\frac{2}{3}\) i.e. at point (0, \(\frac{2}{3}\)).
∴ gradient of the curve at (0, \(\frac{2}{3}\))
= \(\frac{d y}{d x}\) (0, \(\frac{2}{3}\))
= \(-\frac{4 × 0 + 3}{3}\) = 1
Given curve meets x-axis at y = 0 i.e. (2 x + 1)(2 – x) = 0
⇒ x = -\frac{1}{2}, 2
∴ required area = \(\frac{1}{3}\) \(\int_{-1 / 2}^2\) (2 x + 1)(2 – x) d x
= \(\frac{1}{3}\) \(\int_{-1 / 2}^2\) (- 2 x² + 3 x + 2) d x
= \(\frac{1}{3}\)[\(-\frac{2}{3}\) x³ + \(\frac{3 x^2}{2}\) + 2 x]^-1/2 ₂
= \(\frac{1}{3}\)[(\(-\frac{16}{3}\) + 6 + 4) – (\(\frac{1}{12}\) + \(\frac{3}{8}\) – 1)]
= \(\frac{1}{3}\) [\(\frac{14}{3}\) – \(\frac{1}{12}\) – \(\frac{3}{8}\) + 1]
= \(\frac{125}{72}\) sq. units

Question 15.
Given alongside figure shows a sketch of the curve y = (x – 1)(4 – x). If P is the point (2, 2), verify that the tangent at P passes through the origin O as shown. Calculate the area O A P enclosed between the tangent, the curve and the axis of x.

Answer:
The eqn. of given curve be
y = (x – 1)(4 – x)
∴ \(\frac{d y}{d x}\) = (x – 1)(-1) + (4 – x) 1
= -x + 1 + 4 – x = 5 – 2 x

∴ slope of tangent to given curve (1) at point (2, 2)
= \(\frac{d y}{d x}\) {(2,2)} = 5 – 4 = 1
Thus eqn. of tangent through the point (2, 2) be given by
y – 2 = 1(x – 2)
⇒ y = x
Clearly eqn. (2) passes through origin O(0, 0).
∴ area of shaded region OAPO
= area of ∆ OPQ – area of region APQA
= \(\frac{1}{2}\) × 2 × 2 – \(\int_1^2\) (x – 1)(4 – x) d x
[since curve (1) meets x-axis i.e. y = 0 ∴ x = 1, x = 4 ]
= 2 – \(\int_1^2\) (-x² + 5 x – 4) d x
= 2 + \(\int_1^2\) (x² – 5 x + 4) d x
= 2 + \(\frac{x^3}{3}\) – \(\frac{5 x^2}{2}\) + 4 x]²¹
= 2 + \(\frac{8}{3}\) – 10 + 8 – \(\frac{1}{3}\) + \(\frac{5}{2}\) – 4]
= \(\frac{8}{3}\) – \(\frac{1}{3}\) + \(\frac{5}{2}\) – 4
= \(\frac{7}{3}\) – \(\frac{3}{2}\)
= \(\frac{5}{6}\) sq. units

Question 16.
The curve y = a x² + b x + c passes through the points (1, 0),(2, 0) and its gradient at the point (2, 0) is 2 . Find the numerical value of the area included between the curve and the axis of x.
Answer:
The eqn. of given curve y = a x² + b x + c eqn. (1) passes through the point (1, 0) and (2, 0).
∴ 0 = a + b + c
0 = 4 a + 2 b + c
Diff. eqn. (1) both sides w.r.t. x; ; we have
\(\frac{d y}{d x}\) = 2 a x + b
∴ gradient of curve at point (2, 0) = 2
⇒ ( \(\frac{d y}{d x}\)) (2,0) = 2
⇒ 4 a + b = 2
eqn. (3) -2 x eqn. (4) ; we have
4 a + 2 b + c – 8 a – 2 b + 4 = 0
⇒ -4 a + c = -4
⇒ 4 a – c = 4
eqn. (4) – eqn. (2) ; we have
3a – c = 2
eqn. (5) – eqn. (6) gives; a = 2
∴ from (6) ; c = 6 – 2 = 4
∴ from (4); b = 2 – 4 a = 2 – 8 = -6
Thus from (1); we have
y = 2 x² – 6 x + 4
eqn. (7) meets x-axis when y = 0
∴ 2 x² – 6 x + 4 = 0
⇒ (x – 1)(2 x – 4) = 0
⇒ x = 1,2
∴ required area = \(\int_1^2\) y d x
= \(\int_1^2\) (2 x² – 6 x + 4) d x
= 2 \(\int_1^2\) (x² – 3 x + 2) d x
= 2\(\frac{x^3}{3}\) – \(\frac{3 x^2}{2}\) + 2 x]² ₁
= |2\(\frac{8}{3}\) – 6 + 4 – \(\frac{1}{3}\) + \(\frac{3}{2}\) – 2]|
= 2[\(\frac{7}{3}\) + \(\frac{3}{2}\) – 4]
= |2\(\frac{14+9-24}{6}\)|
= \(\frac{1}{3}\) sq. units

Question 17.
The line y = 2 x meets the curve y² = 4 x at the point O (the origin) and P, and P N is perpendicular to the y-axis. Prove that the area between the curve and O P is one-half the area enclosed by the lines O N, N P and the curve.
Answer:
eqn. of given line be
y = 2 x

y² = 4 x
and eqn. of given curve be
eqn. (2) represents a right handed parabola with vertex at (0, 0) and line (1) meets curve (2) when 4 x² = 4 x
⇒ 4 x(x – 1) = 0
⇒ x = 0,1
∴ from (1); y = 0, 2
Thus the point of intersection are O(0, 0) and P(1, 2).
∴ area betwen the curve and OP
A₁ = \(\int_0^1\) [2 \(\sqrt{x}\) – 2 x] d x
= \(\frac{2 x^{3 / 2}}{3 / 2}-x^2\)]_0^1
= \(\frac{4}{3}\) – 1
= \(\frac{1}{3}\) sq. units
A₂ = area of region ONPO
= area of ∆ ONP – area between curve and OP
= \(\frac{1}{2}\) × 2 × 1 – \(\frac{1}{3}\)
= \(\frac{2}{3}\) sq. units
∴ ⇒ \(\frac{\mathrm{A}_1}{\mathrm{~A}_2}\)
= \(\frac{1}{2}\)
⇒ area between the curve and OP is one-half the area enclosed by the lines ON, NP and the curve.

Question 18.
Calculate the areas of the two parts into which the area enclosed by the x-axis and the curve y=18 x-3 x^2 is divided by the line x=4.
Answer:
Given eqn. of curve be
y = 18 x – 3 x²
⇒ y = -3(x² – 6 x + 9 – 9)
⇒ y = -3(x – 3)² + 27
⇒ 3(x – 3)² = 27 – y
⇒ (x – 3)²
= \(-\frac{1}{3}\)(y – 27)
which represents a downward parabola with vertex (3, 27). Curve (1) meets x-axis at y = 0
∴18 x – 3 x² =0
⇒ 3 x(6 – x) = 0
⇒ x = 0, 6
i.e. at points (0, 0) and Q(6, 0)
The line x = 4 meets curve (1)
When y = 72 – 48 = 24 i.e. at P(4, 24)

A₁ = area of shaded region OAPRO
= \(\int_0^4\) y d x
= \(\int_0^4\) (18 x – 3 x² ) d x
= 9 x² – x³]⁰ ₄
= 144 – 64
= 80 sq. units
A₂ = area of shaded region PQRP
= \(\int_4^6\) (18 x – 3 x²) d x
= 9 x² – x³]⁴₆
= (9 × 36 – 216) – (144 – 64) = 108 – 80
= 28 sq. units

Question 19.
Draw the rough sketch of y² + 1 = x, x < 2 and find the area enclosed by the curve and the line x = 2.
Answer:
Given equation of curve be y² + 1 = x
⇒ y² = x – 1 Clearly it represents a parabola (right handed) with vertex (1, 0) and does not meeting y-axis at any point.
Further the given curve meets the line x = 2 at y² + 1 = 2
⇒ y = ± 1 i.e. at points (2, + 1) and (2, – 1).

Clearly the given curve is symmetrical about x-axis.
∴ required area = 2 × area of region enclosed by parabola and line x = 2 in first quadrant.
Divide this region R into vertical strips with lower end on x axis and upper end on y = \(\sqrt{x-1}\) and corresponding rectangle move from x = 1 and x = 2
∴ required area = 2 \(\int_1^2\) |y| d y = 2 \(\int_1^2\) y d x
= 2 \(\int_1^2\) \(\sqrt{x-1}\) d x
= 2 \(\frac{(x-1)^{3 / 2}}{3 / 2}\)]²₁
= \(\frac{4}{3}\)[1 – 0]
= \(\frac{4}{3}\) square units.
[∵ y ≥ 0 ∴ |y| = y]

Question 20.
Find the area bounded by the curve y = 4 – x² and the line y = 0 and y = 3.
Answer:

We want to find the area of region bounded by curves given below:
and
y = 4 – x²
y = 0, y = 3
eqn. (1) represents a parabola (downward) with vertex (0,4). y = 4 – x² meets y = 3 when 3 = 4 – x²
⇒ x² = 1
⇒ x = ± 1
∴ points of intersection are ( ± 1,3).
∴ required area = 2 × area of region OABCO
Divide the region into small horizontal strips. Each strip has left end on y-axis and right end on curve y = 4 – x².
∴ required area = 2 \(\int_0^3\) x d y
= 2 \(\int_0^3\) \(\sqrt{4-y}\) d y
= 2 \(\frac{(4-y)^{3 / 2}}{-3 / 2}\)]⁰ ₃
= \(-\frac{4}{3}\)[1 – 8]
= \(\frac{28}{3}\) sq. units

Question 21.
Make a rough sketch of the following curves. Also, find the area enclosed between the curves and the axes.
(i) y = cos x, 0 ≤ x ≤ \(\frac{\pi}{2}\)
(ii) y = cos 2 x, 0 ≤ x ≤ \(\frac{\pi}{4}\)
(iii) y = sin x, 0 ≤ x ≤ \(\frac{\pi}{2}\)
(iv) y = cos ^2 x, 0 ≤ x ≤ \(\frac{\pi}{2}\)
Answer:
(i) Given eqn. of curve be y = cos x ; 0 ≤ x ≤ \(\frac{\pi}{2}\)
Table of values is given as under:

∴ required area = \(\int_0^{\pi / 2}\) y d x
= \(\int_0^{\pi / 2}\) cos x d x
= sin x]^π/2 ₀
= sin \(\frac{\pi}{2}\) – sin 0
= 1 – 0 = 1 sq. units

(ii) Given eqn. of curve be
y = cos 2 x ; 0 ≤ x ≤ \(\frac{\pi}{4}\)

∴ required area = \(\int_0^{\pi / 4}\) y d x
= \(\int_0^{\pi / 4}\) cos 2 x d x
= \(\frac{\sin 2 x}{2}\)]^π/4 ₀
= \(\frac{1}{2}\) [sin \(\frac{\pi}{2}\) – 0]
= \(\frac{1}{2}\) sq. units

(iii) Equation of given curve be
y = sin x ; 0 ≤ x ≤ \(\frac{\pi}{2}\)

∴ required area = \(\int_0^{\pi / 2}\) y dx = \(\int_0^{\pi / 2}\) sin x d x
= -cos x]^π/2 ₀
= -[0 – 1]
= 1 sq. units

(iv) Given curve is y = cos² x
For rough sketch we construct a table of values as under :

∴ Required area = \(\int_0^{\pi / 2}\) y d x [using vertical strips]
= \(\int_0^{\pi / 2}\) cos² x d x
= \(\frac{1}{2}\) \(\int_0^{\pi / 2}\) (1 + cos 2 x) d x
= \(\frac{1}{2}\) [x + \(\frac{\sin 2 x}{2}\)]^π/2 ₀
= \(\frac{\pi}{4}\) sq. units

Question 22.
Draw a rough sketch of the curve y = cos² x in [0, π] and find the area enclosed by the the lines x = 0, x = π and the x-axis.
Answer:
Given eqn. of curve be y = cos² x in [0, π]

∴ Required area = \(\int_0^{\pi / 2}\) cos² x + \(\int_{\pi / 2}^\pi\) cos² x d x
= \(\int_0^\pi \) cos² x d x
= \(\int_0^\pi\) \(\frac{1+\cos 2 x}{2}\) d x
= \(\frac{1}{2}\) [x + \(\frac{\sin 2 x}{2}\)]^π ₀
= \(\frac{1}{2}\)[π + 0 – 0 – 0]
= \(\frac{\pi}{2}\) sq. units

Question 23.
Draw a rough sketch of the curve y = \(\frac{1}{2}\)π + 2 sin² x and find the area between the x-axis, the curve and the ordinates x = 0 and x = π
Answer:
Given eqn. of curve be y = \(\frac{\pi}{2}\) + 2 sin² x

The rough sketch for given curve is given as under.
Divide the region into vertical strips. Each vertical strip has lower end on x-axis and upper end on given curve.
Thus the corresponding rectangle has length |y| and width d x and this rectangle move from x = 0 and x = π.
∴ Required area = \(\int_0^\pi\) |y| d x = \(\int_0^\pi\) y d x
[∵ y ≥ 0 ∴ |y| = y]
= \(\int_0^\pi\) \(\frac{\pi}{2}\) + 2 sin² x] d x
= \(\frac{\pi}{2}\) x]^π ₀ + \(\int_0^\pi \) 2[ \(\frac{1-\cos 2 x}{2}\)] d x
= \(\frac{\pi}{2}\)(π – 0) + [x – \(\frac{\sin 2 x}{2}\)]^π ₀
= \(\frac{\pi^2}{2}\) + [π – 0 – 0 – 0]
= \(\frac{\pi^2}{2}\) + π
= \(\frac{\pi}{2}\) (π + 2) sq. units

Question 24.
Show that the area included between the x-axis and the curve a² y = x² (x + a) is \(\frac{a^2}{12}\)
Answer:
Clearly the curve a² y = x² (x + a)
meets x-axis i.e. y = 0, where x² (x + a) = 0
⇒ x = 0,-a
∴ curve meets x-axis at (0,0) and (-a, 0).
∴ required area
= \(\int_{-a}^0\) y d x
= \(\int_{-a}^0\) \(\frac{1}{a^2}\) (x³ + a x²) d x
= \(\frac{1}{a^2}\) \(\frac{x^4}{4}\) + \(\frac{a x^3}{3}\)]⁰ _a
= \(\frac{1}{a^2}\) [0 + 0 – \(\frac{a^4}{4}\) + \(\frac{a^4}{3}\)]
= \(\frac{a^2}{12}\) sq. units

Question 24.
Show that the area included between the x-axis and the curve a² y = x² (x + a) is \(\frac{a^2}{12}\)
Answer:
Clearly the curve a² y = x² (x + a)
meets x-axis i.e. y = 0, where x² (x + a) = 0
⇒ x = 0,-a
∴ curve meets x-axis at (0,0) and (-a, 0).
∴ required area
= \(\int_{-a}^0\) y d x
= \(\int_{-a}^0\) \(\frac{1}{a^2}\) (x³ + a x²) d x
= \(\frac{1}{a^2}\) \(\frac{x^4}{4}\) + \(\frac{a x^3}{3}\)]⁰ _a
= \(\frac{1}{a^2}\) [0 + 0 – \(\frac{a^4}{4}\) + \(\frac{a^4}{3}\)]
= \(\frac{a^2}{12}\) sq. units

Question 25.
(i) Calculate the area bounded by the curve y = x² – 1, the x-axis and the line y = 8.
(ii). Calculate the approximate increase in this area, if the line y = 8 is changed to y = 8.01.
Answer:
(i) The given curve y = x² – 1
⇒ x² = y + 1
it represents an upward parabola with vertex (0, -1) and meets x-axis at y = 0.
∴ from (1), x²
= 1 ⇒ x = ± 1
i.e. at points ( ± 1, 0).
The line y = 8 meets the curve (1) when x² = 9
⇒ x = ± 3 i.e. at points ( ± 3, 8).
∴ Required area = 2 \(\int_0^8\) x d y
= 2 \(\int_0^8\) \(\sqrt{y+1}\) d y
= 2 \(\frac{(y+1)^{3 / 2}}{3 / 2}\)]⁰ ₈
= \(\frac{4}{3}\) [9^3/2 – 1]
= \(\frac{4}{3}\) × 26
= \(\frac{104}{3}\) sq. units

(ii) Thus required area = 2 \(\int_8^{8.01}\) x d y
= 2 \(\int_8^{8.01}\) \(\sqrt{y+1}\) d y
= 2 \(\frac{(y+1)^{3 / 2}}{3 / 2}\)]⁸ _0.88
= \(\frac{4}{3}\)[(9.01)^3/2 – 9^3/2 ]
= \(\frac{4}{3}\)[27.0450 – 27]
= 0.06 sq. units

Question 26.
Find the area of the region bounded by y = -1, y = 2, x = y³ and x = 0.
Answer:
The given curve be x = y³ and lines are y = -1 ; y = 2 and x = 0
The line y = 2 intersects the curve x = y³ at (8, 2)
and y = -1 meets curve at the point (-1, -1).
∴ Required area = \(\int_0^2\) x d y + \(\int_{-1}^0\) x d y
= \(\int_0^2 y^3\) d y + \(\int_{-1}^0\) (0 – y³) d y
= \(\frac{y^4}{4}\)]_0^2 – \(\frac{y^4}{4}\)]^-1 2
= 4 – \(\frac{1}{4}\) [0 – 1]
= 4 + \(\frac{1}{4}\)
= \(\frac{19}{4}\) sq. units

Question 27.
Find the area bounded by x = a t² , y = 2 a t between the ordinates corresponding to t = 1 and t = 2.
Answer:
Equation of given curve be x = a t² ; y = 2 a t
On eliminating t, we have y² = 4 a x which is a right handed parabola with vertex (0, 0).
When t = 1 ⇒ x = a, y = 2 a
When t = 2 ⇒ x = 4 a, y = 4 a
∴ required area = 2 × area of shaded region in Ist qudrant. Divide this region into vertical strips. Each vertical strip has lower end on x-axis and upper end on given curve. So the corresponding rectangle has length |y| and width d x and this rectangle move from x = a to x = 4 a.
∴ Required area = 2 \(\int_a^{4 a}\)|y| d x
= 2 \(\int_a^{4 a}\) 2 \(\sqrt{a x}\) d x [∵ y ≥ 0]
= 4 \(\sqrt{a}\) \(\frac{x^{3 / 2}}{3 / 2}\)]^a _4a
= \(\frac{8 \sqrt{a}}{3}\) [(4 a)^3/2 – a^3/2 ]
= \(\frac{8 \sqrt{a}}{3}\)[8 – 1] a^3/2
= \(\frac{56}{3}\) a² sq. units

Students often turn to ISC S Chand Maths Class 12 Solutions Chapter 6 Matrices Ex 6(i) to clarify doubts and improve problem-solving skills.

S Chand Class 12 ICSE Maths Solutions Chapter 6 Matrices Ex 6(i)

Using elementary transformation, find the inverse of the following matrices, if it exists.

Question 1.
A = \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
Since A = IA ⇒ \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)A
Operate R₂ → R₂ – 2R₁
\(\left[\begin{array}{ll}
1 & -1 \\
0 & -5
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
-2 & 1
\end{array}\right]\)A
R₂ → \(\frac { 1 }{ 5 }\)R₂
\(\left[\begin{array}{cc}
1 & -1 \\
0 & 0
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
-2 / 5 & 1 / 5
\end{array}\right]\)A
Operate R₂ → R₁ + R₂
\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
3 / 5 & 1 / 5 \\
-2 / 5 & 1 / 5
\end{array}\right]\)A
∴ A^-1 = \(\left[\begin{array}{cc}
3 / 5 & 1 / 5 \\
-2 / 5 & 1 / 5
\end{array}\right]\)
[using def. of inverse, A^-1A = I]

Question 2.
A = \(\left[\begin{array}{rr}
1 & 2 \\
2 & -1
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{rr}
1 & 2 \\
2 & -1
\end{array}\right]\)
Since A = IA ⇒ \(\left[\begin{array}{rr}
1 & 2 \\
2 & -1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)A
Operate R₂ → R₂ – 2R₁
\(\left[\begin{array}{cc}
1 & 2 \\
0 & -5
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
-2 & 1
\end{array}\right]\)A
Operate R₂ → \(\frac { -1 }{ 5 }\)R₂
\(\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
2 / 5 & -1 / 5
\end{array}\right]\)A
Operate R₁ → R₁ – 2R₂
\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 / 5 & 2 / 5 \\
2 / 5 & -1 / 5
\end{array}\right]\)A
∴ A^-1 = \(\left[\begin{array}{cc}
1 & 2 / 5 \\
2 / 5 & -1 / 5
\end{array}\right]\)
[using def. of inverse, A^-1A = I]

Question 3.
\(\left[\begin{array}{ll}
9 & 5 \\
7 & 4
\end{array}\right]\)
Solution:

Question 4.
\(\left[\begin{array}{rr}
10 & -2 \\
-5 & 1
\end{array}\right]\)
Solution:
\(\left[\begin{array}{rr}
10 & -2 \\
-5 & 1
\end{array}\right]\)
Since A = IA ⇒ \(\left[\begin{array}{rr}
10 & -2 \\
-5 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)A
Operate R₁ → R₁ – 2R₂
\(\left[\begin{array}{cc}
0 & 0 \\
-5 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right]\)A
Since these are all zeros in R₁ of left side matrix
Thus A^-1 does not exists.

Question 5.
\(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)
Solution:

Question 6.
\(\left[\begin{array}{rrr}
1 & -1 & 0 \\
2 & 5 & 3 \\
0 & 2 & 1
\end{array}\right]\)
Solution:

Question 7.
\(\left[\begin{array}{rrr}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)
Solution:

Accessing ISC S Chand Maths Class 12 Solutions Chapter 6 Matrices Ex 6(h) can be a valuable tool for students seeking extra practice.

S Chand Class 12 ICSE Maths Solutions Chapter 6 Matrices Ex 6(h)

Question 1.
Show that A is a singular matrix if
(i) A = \(\left[\begin{array}{ll}
3 & 6 \\
2 & 4
\end{array}\right]\)
(ii) A = \(\left[\begin{array}{rrr}
1 & 1 & 3 \\
2 & 2 & 6 \\
2 & -3 & 1
\end{array}\right]\)
Solution:
(i) Given A = \(\left[\begin{array}{ll}
3 & 6 \\
2 & 4
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ll}
3 & 6 \\
2 & 4
\end{array}\right|\) = 12 – 12 = 0
∴ A be a singular matrix

(ii) Given A = \(\left[\begin{array}{rrr}
1 & 1 & 3 \\
2 & 2 & 6 \\
2 & -3 & 1
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{rrr}
1 & 1 & 3 \\
2 & 2 & 6 \\
2 & -3 & 1
\end{array}\right]\)
Expanding along R₁
= 1(2 + 18) – 1(2 – 12) + 3(- 6 – 4)
= 20 + 10 – 30 = 0
Thus A be a singular matrix.

Question 2.
Find x if \(\left[\begin{array}{rrr}
8 & -6 & 2 \\
-6 & 7 & -4 \\
2 & -4 & x
\end{array}\right]\) is a singular matrix.
Solution:
Let A = \(\left[\begin{array}{rrr}
8 & -6 & 2 \\
-6 & 7 & -4 \\
2 & -4 & x
\end{array}\right]\)
since A be a singular matrix ∴ |A| = 0
\(\left[\begin{array}{rrr}
8 & -6 & 2 \\
-6 & 7 & -4 \\
2 & -4 & x
\end{array}\right]\) = 0; Expanding along R₁
⇒ 8(7x – 16) + 6(-6x + 8) + 2(24 – 14) = 0
⇒ 56x – 128 – 36x + 48 + 48 – 28 = 0
⇒ 20x – 60 = 0
⇒ x = 3

Question 3.
For each of the following matrices, determine wheter the inverse exists. If it exists, find it.
(i) \(\left[\begin{array}{ll}
4 & 5 \\
2 & 3
\end{array}\right]\)
(ii) \(\left[\begin{array}{rr}
2 & -6 \\
-1 & 3
\end{array}\right]\)
(iii) \(\left[\begin{array}{ll}
2 & 6 \\
1 & 1
\end{array}\right]\)
(iv) \(\left[\begin{array}{rr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\)
Solution:
(i) Let A = \(\left[\begin{array}{ll}
4 & 5 \\
2 & 3
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
4 & 5 \\
2 & 3
\end{array}\right|\) = 12 – 10 = 2 ≠ 0
Thus A^-1 exists ∴ A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
∴ The cofactors of R₁ are ; 3 ; – 2
The cofactors of R₂ are ; – 5 ; 4
∴ adj A = \(\left[\begin{array}{cc}
3 & -2 \\
-5 & 4
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
3 & -5 \\
-2 & 4
\end{array}\right]\)
Thus A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A = \(\frac{1}{2}\left[\begin{array}{cc}
3 & -5 \\
-2 & 4
\end{array}\right]\)

(ii) Let A = \(\left[\begin{array}{rr}
2 & -6 \\
-1 & 3
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{rr}
2 & -6 \\
-1 & 3
\end{array}\right]\)
Thus A is a singular matrix
∴ A^-1 does not exists.

(iii) Let A = \(\left[\begin{array}{ll}
2 & 6 \\
1 & 1
\end{array}\right]\);
Here |A| = \(\left|\begin{array}{ll}
2 & 6 \\
1 & 1
\end{array}\right|\) = 2 – 6 = – 4 ≠ 0
Thus A^-1 exists
∴ A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A … (i)
The cofactor of R₁ are ; 1 ; – 1
The cofactor of R₂ are ; – 6; 2
∴ adj A = \(\left[\begin{array}{cc}
1 & -1 \\
-6 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
1 & -6 \\
-1 & 2
\end{array}\right]\)
Thus from (1) ; we have
A^-1 = \(\frac{1}{-4}\left[\begin{array}{cc}
1 & -6 \\
-1 & 2
\end{array}\right]=\frac{1}{4}\left[\begin{array}{cc}
-1 & 6 \\
1 & -2
\end{array}\right]\)

(iv) Let A = \(\left[\begin{array}{rr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\)

Question 4.
(a) If A = \(\left[\begin{array}{ll}
2 & x \\
4 & 2
\end{array}\right]\), x ≠ 1, calculate
(i) A²
(ii) (A²)^-1
(b) Find the sum of \(\left[\begin{array}{ll}
2 & 3 \\
5 & 7
\end{array}\right]\) multiplicative inverse.
Solution:
(a) Given A = \(\left[\begin{array}{ll}
2 & x \\
4 & 2
\end{array}\right]\)
∴ A² = A.A = \(\left[\begin{array}{ll}
2 & x \\
4 & 2
\end{array}\right]\left[\begin{array}{ll}
2 & x \\
4 & 2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
4+4 x & 2 x+2 x \\
8+8 & 4 x+4
\end{array}\right]\)
∴ A² = \(\left[\begin{array}{cc}
4+4 x & 4 x \\
16 & 4 x+4
\end{array}\right]\)
∴ |B| = \(\left|\begin{array}{cc}
4+4 x & 4 x \\
16 & 4 x+4
\end{array}\right|\)
= (4 + 4x)² – 64x = (4 – 4x)²
= 16(1 – x)²
= 16(x – 1)² ≠ 0 [ ∵ x ≠ 1]
The cofactor of R₁ are ; + 4x + 4 ; – 16
The cofactor of R₂ are ; – 4x ; 4x + 4

The cofactor of R₁ are ; 7 ; -5
The cofactor of R₂ are ; – 3 ; 2

Question 5.
(i) Find a matrix X for which
X\(\left[\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right]=\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right]\). Also find the inverse of \(\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right]\).
(ii) Find 2 x 2 matrix M such that
\(\left[\begin{array}{rr}
-5 & 2 \\
15 & -7
\end{array}\right] M=\left[\begin{array}{ll}
5 & 0 \\
0 & 5
\end{array}\right]\).
Solution:
Given X\(\left[\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right]=\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right]\)
∴ X = \(\left[\begin{array}{rr}
4 & 1 \\
2 & 3
\end{array}\right]\left[\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right]^{-1}\) … (1)
Let A = \(\left[\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right]\)
∴\(|\mathrm{A}|=\left|\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right|\)
= – 3 – 2 = – 5 ≠ 0
Thus A^-1 exists
∴ A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
The cofactor of R₁ are ; – 1 ; – 1
The cofactor of R₂ are ; – 2 ; 3

Thus B^-1 exists & B^-1 = \(\frac{1}{|\mathrm{~B}|}\) adj B
The cofactor of R₁ are ; 3 ; – 2
The cofactor of R₂ are ; – 1 ; 4

The cofactor of R₁ are ; -7 ; -15
The cofactor of R₂ are ; -2 ; -15

Question 6.
Find the inverse of the following matrices and verify your result.
(i) \(\left[\begin{array}{rr}
2 & 5 \\
-3 & 1
\end{array}\right]\)
(ii) \(\left[\begin{array}{rr}
-4 & 3 \\
5 & -5
\end{array}\right]\)
Solution:
(i) Let A = \(\left[\begin{array}{rr}
2 & 5 \\
-3 & 1
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{rr}
2 & 5 \\
-3 & 1
\end{array}\right|\)
Thus, A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
The cofactor of R₁ are ; 1 ; 3
The cofactor of R₂ are ; – 5 ; 2

Question 7.
(a) Find the inverse of each of the following matrices and verify your result.
(i) \(\left[\begin{array}{rrr}
1 & 4 & 3 \\
2 & 5 & 4 \\
1 & -3 & -2
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccr}
1 & 0 & 0 \\
0 & \cos \alpha & \sin \alpha \\
0 & \sin a & -\cos \alpha
\end{array}\right]r\)
(b) Verify that AA^-1 = A^-1A = I, if
\(\left[\begin{array}{rrr}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]\)
(c) If A = \(\left[\begin{array}{rrr}
3 & 0 & 2 \\
1 & 5 & 9 \\
-6 & 4 & 7
\end{array}\right]\) and AB = BA = I, find B.
Solution:

(ii) Let A = \(\left[\begin{array}{ccr}
1 & 0 & 0 \\
0 & \cos \alpha & \sin \alpha \\
0 & \sin a & -\cos \alpha
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \alpha & \sin \alpha \\
0 & \sin \alpha & -\cos \alpha
\end{array}\right|\);
Expanding along R₁ = – cos²α – sin²α = – 1 ≠ 0
∴ A^-1 exists
Thus, A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A … (1)
The cofactor of R₁ are ;

Expanding along C₁
= 1(3 – 0) +1(2 – 4) = 3 – 2 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
The cofactor of R₁ are ;
\(\left|\begin{array}{rr}
3 & 0 \\
-2 & -1
\end{array}\right| ;-\left|\begin{array}{rr}
-1 & 0 \\
0 & 1
\end{array}\right| ;\left|\begin{array}{rr}
-1 & 3 \\
0 & -2
\end{array}\right|\)
i.e. 3 ; 1 ; 2
The cofactor of R₂ are ;
\(-\left|\begin{array}{rr}
2 & -2 \\
-2 & 1
\end{array}\right| ;\left|\begin{array}{rr}
1 & -2 \\
0 & 1
\end{array}\right| ;-\left|\begin{array}{rr}
1 & 2 \\
0 & -2
\end{array}\right|\)
i.e. 2 ; 1 ; 2
The cofactor of R₃ are ;

(c) Let A = \(\left[\begin{array}{rrr}
3 & 0 & 2 \\
1 & 5 & 9 \\
-6 & 4 & 7
\end{array}\right]\)
since AB = BA = I,
Also, AA’ = A^-1A = I
∴ B = A^-1
Now, |A| = \(\left|\begin{array}{rrr}
3 & 0 & 2 \\
1 & 5 & 9 \\
-6 & 4 & 7
\end{array}\right|\)
Expanding along R₁
= 3(35 – 36) + 0 + 2(4 + 30)
= – 3 + 68 = 65 ≠ 0
∴ A^-1 exists
& A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
The cofactor of R₁ are ;
\(\left|\begin{array}{ll}
5 & 9 \\
4 & 7
\end{array}\right| ;-\left|\begin{array}{cc}
1 & 9 \\
-6 & 7
\end{array}\right| ;\left|\begin{array}{cc}
1 & 5 \\
-6 & 4
\end{array}\right|\)
The cofactor of R₂ are
\(-\left|\begin{array}{cc}
0 & 2 \\
4 & 7
\end{array}\right| ;\left|\begin{array}{cc}
3 & 2 \\
-6 & 7
\end{array}\right| ;-\left|\begin{array}{cc}
3 & 0 \\
-6 & 4
\end{array}\right|\)
i.e. 8 ; 33 ; – 12
The cofactor of R₃ are
\(\left|\begin{array}{ll}
0 & 2 \\
5 & 9
\end{array}\right| ;-\left|\begin{array}{ll}
3 & 2 \\
1 & 9
\end{array}\right| ;\left|\begin{array}{ll}
3 & 0 \\
1 & 5
\end{array}\right|\)
i.e. -10 ; -25 ; 15
∴ adj A = \(\left[\begin{array}{ccc}
-1 & -61 & 34 \\
8 & 33 & -12 \\
-10 & -25 & 15
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-1 & 8 & -10 \\
-61 & 33 & -25 \\
34 & -12 & 15
\end{array}\right]\)
Thus, B = A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
= \(\frac{1}{65}\left[\begin{array}{ccc}
-1 & 8 & -10 \\
-61 & 33 & -25 \\
34 & -12 & 15
\end{array}\right]\)

Question 8.
Verify that (AB)^-1 = B^-1A^-1 for the matrices A and B where
(i) A = \(\left[\begin{array}{ll}
2 & 1 \\
5 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
4 & 5 \\
3 & 4
\end{array}\right]\)
(ii) A = \(\left[\begin{array}{ll}
3 & 2 \\
7 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
4 & 6 \\
3 & 2
\end{array}\right]\)
Solution:
(i) Given A = \(\left[\begin{array}{ll}
2 & 1 \\
5 & 3
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{ll}
2 & 1 \\
5 & 3
\end{array}\right]\) = 6 – 5 = 1 ≠ 0
Thus, A^-1 exists and A^-1 =\(\frac{1}{|\mathrm{~A}|}\) adj A …(1)
The cofactor of R₁ are ; 3 ; – 5
The cofactor of R₂ are ; – 1 ; 2
∴ adj A = \(\left[\begin{array}{cc}
3 & -5 \\
-1 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
3 & -1 \\
-5 & 2
\end{array}\right]\)
Thus, from (1);
A^-1 = \(\frac{1}{1}\left[\begin{array}{cc}
3 & -1 \\
-5 & 2
\end{array}\right]=\left[\begin{array}{cc}
3 & -1 \\
-5 & 2
\end{array}\right]\)
& B = \(\left[\begin{array}{ll}
4 & 5 \\
3 & 4
\end{array}\right]\)
∴ |B| = \(\left|\begin{array}{ll}
4 & 5 \\
3 & 4
\end{array}\right|\)
= 16 – 15 = 1 ≠ 0
Thus, B^-1 exists and B^-1 =\(\frac{1}{|\mathrm{~A}|}\) adj A …(2)
The cofactor of R₁ are ; 4 ; – 3
The cofactor of R₂ are ; – 5 ; 4

The cofactor of R₁ are ; 37 ; – 29
The cofactor of R₂ are ; – 14 ; 11
∴ adj (AB)
= \(\left[\begin{array}{cc}
37 & -29 \\
-14 & 11
\end{array}\right]=\left[\begin{array}{cc}
37 & -14 \\
-29 & 11
\end{array}\right]\)
Thus (AB)^-1 = \(\frac{1}{|\mathrm{AB}|}\)adj (AB)
= \(\left[\begin{array}{cc}
37 & -14 \\
-29 & 11
\end{array}\right]\) … (2)
[ ∵ |AB| = \(\left|\begin{array}{ll}
11 & 14 \\
29 & 37
\end{array}\right|=407-406=1 \neq 0\)
∴ (AB)^-1 exists]
from (1) & (2); we have
B^-1A^-1 = (AB)^-1

(ii) Given A = \(\left[\begin{array}{ll}
3 & 2 \\
7 & 5
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
3 & 2 \\
7 & 5
\end{array}\right|=15-14=1 \neq 0\)
∴ A^-1 exists & A^-1 = \(\frac { 1 }{ |A| }\)adj A
The cofactor of R₁ are ; 5 ; – 7
The cofactor of R₂ are ; – 2 ; 3

The cofactor of R₁ are ; 2 ; – 3
The cofactor of R₂ are ; – 6 ; 4

The cofactor of R₁ are ; 52 ; – 43
The cofactor of R₂ are ; – 22 ; 18
∴ adj (AB)
= \(\left[\begin{array}{cc}
52 & -43 \\
-22 & 18
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
52 & -22 \\
-43 & 18
\end{array}\right]\)
Thus, (AB)^-1 = \(\frac{1}{|\mathrm{AB}|}\)adj (AB)
= \(\frac{-1}{10}\left[\begin{array}{cc}
52 & -22 \\
-43 & 18
\end{array}\right]=\frac{1}{10}\left[\begin{array}{cc}
-52 & 22 \\
43 & -18
\end{array}\right]\)
from (1) & (2) ; we have
(AB)^-1 = B^-1A^-1

Question 9.
Given A = \(\left[\begin{array}{rr}
2 & -3 \\
-4 & 7
\end{array}\right]\), compute A^-1 and show that 2A^-1 = 9I – A.
Solution:
Given A = \(\left[\begin{array}{rr}
2 & -3 \\
-4 & 7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{rr}
2 & -3 \\
-4 & 7
\end{array}\right|\) = 14 – 12 = 2 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac { 1 }{ |A| }\)adj A … (1)
The cofactor of R₁ are ; 7 ; 4
The cofactor of R₂ are ; 3 ; 2