ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.1

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.1

Question 1.
Evaluate the following determinants:
(i) \(\left|\begin{array}{cc}
x^2-x+x & x-1 \\
x+1 & x+1
\end{array}\right|\) (NCERT)
(ii) \(\left|\begin{array}{rr}
\cos 90^{\circ} & -\cos 45^{\circ} \\
\sin 90^{\circ} & \sin 45^{\circ}
\end{array}\right|\)
(iii) \(\left|\begin{array}{rr}
2 \cos \theta & -2 \sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
Solution:
(i) \(\left|\begin{array}{cc}
x^2-x+x & x-1 \\
x+1 & x+1
\end{array}\right|\)
= (x² – x + 1) (x + 1) – (x – 1) (x + 1)
= x³ + 1 – (x² – 1)
= x³ – x² + 2

(ii) \(\left|\begin{array}{rr}
\cos 90^{\circ} & -\cos 45^{\circ} \\
\sin 90^{\circ} & \sin 45^{\circ}
\end{array}\right|\)
= \(0 \times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \times 1\)
= \(\frac{1}{\sqrt{2}}\)

(iii) \(\left|\begin{array}{rr}
2 \cos \theta & -2 \sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
= 2 cos² θ + 2 sin² θ
= 2 (cos² θ + sin² θ)
= 2.

Question 2.
Evaluate the following determinants:
(i) \(\left|\begin{array}{rr}
\sin 30^{\circ} & \cos 30^{\circ} \\
-\sin 60^{\circ} & \cos 60^{\circ}
\end{array}\right|\)
(ii) \(\left|\begin{array}{rrr}
-1 & 31 & 40 \\
0 & 5 & -432 \\
0 & 0 & 20
\end{array}\right|\)
(iii) \(\left|\begin{array}{rrr}
2 & 0 & 0 \\
-25 & 3 & 0 \\
71 & 64 & 7
\end{array}\right|\)
Solution:
(i) \(\left|\begin{array}{rr}
\sin 30^{\circ} & \cos 30^{\circ} \\
-\sin 60^{\circ} & \cos 60^{\circ}
\end{array}\right|\)
= \(\left|\begin{array}{rr}
\frac{1}{2} & \frac{\sqrt{3}}{2} \\
-\frac{\sqrt{3}}{2} & \frac{1}{2}
\end{array}\right|\)
= \(\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\)
= \(\frac{1}{4}+\frac{3}{4}\) = 1

(ii) Let |A| = \(\left|\begin{array}{rrr}
-1 & 31 & 40 \\
0 & 5 & -432 \\
0 & 0 & 20
\end{array}\right|\) ;
expanding along C₁
= – 1 \(\left|\begin{array}{rr}
5 & -432 \\
0 & 20
\end{array}\right|\)
= – 1 (100 + 0)
= – 100

(iii) \(\left|\begin{array}{rrr}
2 & 0 & 0 \\
-25 & 3 & 0 \\
71 & 64 & 7
\end{array}\right|\)
Expanding along R₁
= 2 \(\)
= 2 (21 – 0)
= 42.

Question 3.
(i) Find the value of \(\left|\begin{array}{rr}
\sin A & -\sin B \\
\cos A & \cos B
\end{array}\right|\), where A = 63°, B = 27°.
(ii) Find the value of \(\left|\begin{array}{rr}
\cos A & \sin A \\
-\sin B & \cos B
\end{array}\right|\), where A = 75°, B = 45°.
Solution:
\(\left|\begin{array}{rr}
\sin A & -\sin B \\
\cos A & \cos B
\end{array}\right|\)
= sin A cos B + cos A sin B
= sin (A + B)
= sin (63° + 27°)
= sin 90° = 1.

(ii) \(\left|\begin{array}{rr}
\cos A & \sin A \\
-\sin B & \cos B
\end{array}\right|\)
= cos A cos B + sin A sin B
= cos (A – B)
= cos (75° – 45°)
= cos 30°
= \(\frac{\sqrt{3}}{2}\)

Question 4.
(i) If A = \(\left|\begin{array}{rrr}
5 & 6 & -3 \\
-4 & 3 & 2 \\
-4 & -7 & 3
\end{array}\right|\), then write the cofactor of the element a₂₁ of its 2nd row.
(ii) If A_ij is the cofactor of the element a_ij of the determinant \(\left|\begin{array}{rrr}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right|\), then write the value of a₃₂ A₃₂.
Solution:
(i) Given A =\(\left|\begin{array}{rrr}
5 & 6 & -3 \\
-4 & 3 & 2 \\
-4 & -7 & 3
\end{array}\right|\)
Cofactor of a₂₁ = A₂₁
= (- 1)²⁺¹ \(\left|\begin{array}{rr}
6 & -3 \\
-7 & 3
\end{array}\right|\)
= – (18 – 21)
= 3

(ii) Given |A| = \(\left|\begin{array}{rrr}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right|\)
Here a₃₂ = 5
∴ A₃₂ = cofactor of a₃₂
= (- 1)³⁺² \(\left|\begin{array}{ll}
2 & 5 \\
6 & 4
\end{array}\right|\)
= – (8 – 30) = 22
Thus, a₃₂ A₃₂ = 5 × 22 = 110.

Question 4 (old).
(i) Write the cofactor of a₁₂ in the determinant \(\left|\begin{array}{rrr}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right|\).
Solution:
Given |A| = \(\left|\begin{array}{rrr}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right|\)
∴ Cofactor of a₁₂ = A₁₂
= (- 1)¹⁺²
= \(\left|\begin{array}{rr}
6 & 4 \\
1 & -7
\end{array}\right|\)
= – (42 – 4) = 46.

Question 5.
Find the value of x if
(i) \(\left|\begin{array}{ll}
2 x+5 & 3 \\
5 x+2 & 9
\end{array}\right|\) = 0
(ii) \(\left|\begin{array}{ll}
x+2 & 3 \\
x+5 & 4
\end{array}\right|\) = 3.
(iii) \(\left|\begin{array}{cc}
3 x & 7 \\
-2 & 4
\end{array}\right|=\left|\begin{array}{ll}
8 & 7 \\
6 & 4
\end{array}\right|\) (NCERT)
Solution:
(i) Given \(\left|\begin{array}{ll}
2 x+5 & 3 \\
5 x+2 & 9
\end{array}\right|\) = 0
⇒ 9 (2x + 5) – 3 (5x + 2) = 0
⇒ 18x + 45 – 15x – 10 = 0
⇒ 3x = 39
⇒ x = 13

(ii) \(\left|\begin{array}{ll}
x+2 & 3 \\
x+5 & 4
\end{array}\right|\) = 3
⇒ 4 (x + 2) – 3 (x + 5) = 3
⇒ 4x + 8 – 3x – 15 = 0
⇒ x – 10 = 0
⇒ x = 10.

(iii) Given \(\left|\begin{array}{cc}
3 x & 7 \\
-2 & 4
\end{array}\right|=\left|\begin{array}{ll}
8 & 7 \\
6 & 4
\end{array}\right|\)
⇒ 12x + 14 = 32 – 42
⇒ 12x + 14 = – 10
⇒ 12x = – 24
⇒ x = – 2

Question 5 (old).
(ii) \(\left|\begin{array}{cc}
x & 4 \\
2 & 2 x
\end{array}\right|\) = 0
Solution:
\(\left|\begin{array}{cc}
x & 4 \\
2 & 2 x
\end{array}\right|\) = 0
⇒ x × 2x – 8 = 0
⇒ 2x² – 8 = 0
⇒ x = ± 2.

Question 6.
(i) If \(\left|\begin{array}{ll}
x+1 & x-1 \\
x-3 & x+2
\end{array}\right|=\left|\begin{array}{rr}
4 & -1 \\
1 & 3
\end{array}\right|\), then write the value of x.
(ii) \(\left|\begin{array}{cc}
\sqrt{6} & x \\
\sqrt{20} & \sqrt{24}
\end{array}\right|=\left|\begin{array}{cc}
6 & 2 \\
5 & 2
\end{array}\right|\), then weite the value of x.
Solution:
(i) Given, \(\left|\begin{array}{ll}
x+1 & x-1 \\
x-3 & x+2
\end{array}\right|=\left|\begin{array}{rr}
4 & -1 \\
1 & 3
\end{array}\right|\)
⇒ (x + 1) (x + 2) – (x – 1) (x – 3) = 12 + 1
⇒ (x² + 3x + 2) – (x² – 4x + 3) = 13
⇒ 7x – 1 = 13
⇒ 7x = 14
⇒ x = 2

(ii) Given, \(\left|\begin{array}{cc}
\sqrt{6} & x \\
\sqrt{20} & \sqrt{24}
\end{array}\right|=\left|\begin{array}{cc}
6 & 2 \\
5 & 2
\end{array}\right|\)
⇒ √6 \(\sqrt{24}\) – x \(\sqrt{20}\) = 12 – 10
⇒ \(\sqrt{24 \times 6}\) – \(\sqrt{20}\) × x = 2
⇒ 12 – x \(\sqrt{20}\) = 2
⇒ x \(\sqrt{20}\) = 10
⇒ x = \(\frac{10}{\sqrt{20}}\)
= \(\frac{10}{2 \sqrt{5}}\)
= √5

Question 6 (old).
Find the value of x if
(i) \(\left|\begin{array}{cc}
3 x & 7 \\
-2 & 4
\end{array}\right|=\left|\begin{array}{ll}
8 & 7 \\
6 & 4
\end{array}\right|\)
(ii) \(\left|\begin{array}{cc}
x & 3 \\
2 x & 5
\end{array}\right|=\left|\begin{array}{cc}
2 & 3 \\
4 & 5
\end{array}\right|\) (NCERT)
Solution:
(i) Given \(\left|\begin{array}{cc}
3 x & 7 \\
-2 & 4
\end{array}\right|=\left|\begin{array}{ll}
8 & 7 \\
6 & 4
\end{array}\right|\)
⇒ 3x × 4 – (- 2) × 7 = 8 × 4 – 6 × 7
⇒ 12x + 14 = 32 – 42
⇒ 12x = – 24
⇒ x = – 2

(ii) Given \(\left|\begin{array}{rr}
x & 3 \\
2 x & 5
\end{array}\right|=\left|\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right|\)
⇒ 5x – 6x = 10 – 12
⇒ – x = – 2
⇒ x = 2.

Question 7.
(i) If x ∈ N and \(\left|\begin{array}{rr}
x & 3 \\
4 & x
\end{array}\right|=\left|\begin{array}{rr}
4 & -3 \\
0 & 1
\end{array}\right|\), find the value(s) of x.
(ii) If x ∈ I and \(\left|\begin{array}{cc}
2 x & 3 \\
-1 & x
\end{array}\right|=\left|\begin{array}{cc}
3 & 1 \\
x & 3
\end{array}\right|\), find the value(s) of x.
Solution:
(i) Given \(\left|\begin{array}{rr}
x & 3 \\
4 & x
\end{array}\right|=\left|\begin{array}{rr}
4 & -3 \\
0 & 1
\end{array}\right|\)
⇒ x² – 12 = 4
⇒ x² = 16
⇒ x² = ± 4
But x ∈ N ∴ x = – 4 is rejected.
Thus x = 4.

(ii) Given, \(\left|\begin{array}{cc}
2 x & 3 \\
-1 & x
\end{array}\right|=\left|\begin{array}{cc}
3 & 1 \\
x & 3
\end{array}\right|\)
⇒ 2x² + 3 = 9 – x
⇒ 2x² + x – 6 =0
⇒ 2x² + 4x – 3x – 6 = 0
⇒ 2x (x + 2) – 3 (x + 2) = 0
⇒ (x + 2) (2x – 3) = 0
⇒ x = – 2, 3/2
But x ∈ I, ∴ x = – 2.

Question 8 (old).
(i) If x ∈ N and \(\left|\begin{array}{cc}
x+3 & -2 \\
-3 x & 2 x
\end{array}\right|\) = 8, find the value of x.
(iii) If \(\left|\begin{array}{cc}
x & x \\
1 & x
\end{array}\right|=\left|\begin{array}{ll}
3 & 4 \\
1 & 2
\end{array}\right|\), write the positive value of x.
Solution:
(i) Given \(\left|\begin{array}{cc}
x+3 & -2 \\
-3 x & 2 x
\end{array}\right|\) = 8
⇒ (x + 3) 2x – 6x = 8
⇒ 2x² + 6x – 6x = 8
⇒ 2x² = 8
⇒ x² = 4
⇒ x = ± 2
but x ∈ N
∴ x = – 2 is rejected
Thus x = 2

(iii) Given, \(\left|\begin{array}{cc}
x & x \\
1 & x
\end{array}\right|=\left|\begin{array}{ll}
3 & 4 \\
1 & 2
\end{array}\right|\)
⇒ x² – x = 6 – 4
⇒ x² – x – 2 = 0
⇒ (x – 2) (x + 1) = 0
⇒ x = 2, – 1
Thus required positive value of x be 2.

Question 8.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\), then show that |2A| = 4 |A|. (NCERT)
Solution:
Given A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\)
⇒ |A| = \(\left|\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right|\)
= 2 – 8 = – 6
∴ R.H.S. = 4 |A|
= 4 × (- 6)
= – 24
Now 2A = 2 \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right]\)
∴ |2A| = 8 – 32
= – 24
∴ |2A| = – 24
= L.H.S.
Thus, |2A| = 4 |A|.

Question 9.
(i) \(\left|\begin{array}{rrr}
0 & -1 & 2 \\
1 & 0 & 3 \\
-2 & -3 & 0
\end{array}\right|\)
(ii) \(\left|\begin{array}{rrr}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|\) (NCERT)
(iii) \(\left|\begin{array}{rrr}
2 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right|\) (NCERT)
Solution:
(i) Let |A| = \(\left|\begin{array}{rrr}
0 & -1 & 2 \\
1 & 0 & 3 \\
-2 & -3 & 0
\end{array}\right|\) ;
Expanding along R₁
= 0 + 1 (0 + 6) + 2 (- 3)
= 6 – 6 = 0

Aliter:
Here A’ = \(\left[\begin{array}{rrr}
0 & 1 & -2 \\
-1 & 0 & -3 \\
2 & 3 & 0
\end{array}\right]\)
= \(\left|\begin{array}{rrr}
0 & -1 & 2 \\
1 & 0 & 3 \\
-2 & -3 & 0
\end{array}\right|\)
= – A
∴ A is skew symmetric matrix.
∴ |A| = 0
[∵ A’ = – A
⇒ |A’| = |- A|
⇒ |A| = (- 1)³ |A|
⇒ 2 |A| = 0
⇒ |A| = 0]

(ii) |A| = \(\left|\begin{array}{rrr}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|\) ;
expanding along R₁
= \(3\left|\begin{array}{rr}
1 & -2 \\
3 & 1
\end{array}\right|+4\left|\begin{array}{rr}
1 & -2 \\
2 & 1
\end{array}\right|+5\left|\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right|\)
= 3 (1 + 6) + 4 (1 + 4) + 5 (3 – 2)
= 21 + 20 + 5 = 46

(iii) Let |A| = \(\left|\begin{array}{rrr}
2 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right|\) ;
expanding along R₁
= \(2\left|\begin{array}{rr}
2 & -1 \\
-5 & 0
\end{array}\right|+1\left|\begin{array}{rr}
0 & -1 \\
3 & 0
\end{array}\right|-2\left|\begin{array}{rr}
0 & 2 \\
3 & -5
\end{array}\right|\)
= 2 (0 – 5) + 1 (0 + 3) – 2 (0 – 6)
= – 10 + 3 + 12 = 5

Question 10.
Prove that \(\left|\begin{array}{rrr}
1 & a & b \\
-a & 1 & c \\
-b & -c & 1
\end{array}\right|\) = 1 + a² + b² + c².
Solution:
Let |A| = \(\left|\begin{array}{rrr}
1 & a & b \\
-a & 1 & c \\
-b & -c & 1
\end{array}\right|\) ;
expanding along R₁
= \(1\left|\begin{array}{rr}
1 & c \\
-c & 1
\end{array}\right|-a\left|\begin{array}{rr}
-a & c \\
-b & 1
\end{array}\right|+b\left|\begin{array}{rr}
-a & 1 \\
-b & -c
\end{array}\right|\)
= (1 + c²) – a (- a + bc) + b (ac + b)
= 1 + c² + a² – abc + abc + b²
= 1 + a² + b² + c²

Question 11.
Show that the value of the determinant \(\left|\begin{array}{ccc}
0 & \tan x & 1 \\
1 & -\sec x & 0 \\
\sec x & 0 & \tan x
\end{array}\right|\) is independent of x.
Solution:
Let |A| = \(\left|\begin{array}{ccc}
0 & \tan x & 1 \\
1 & -\sec x & 0 \\
\sec x & 0 & \tan x
\end{array}\right|\) ;
Expanding along R₁
= – tan x \(\left|\begin{array}{cc}
1 & 0 \\
\sec x & \tan x
\end{array}\right|\) + 1 \(\left|\begin{array}{cc}
1 & -\sec x \\
\sec x & 0
\end{array}\right|\)
= – tan x (tan x – 0) + 1 (0 + sec² x)
= – tan² x + sec² x
= 1

Question 11 (old).
(i) \(\left|\begin{array}{rrr}
2 & 4 & 1 \\
8 & 5 & 2 \\
-1 & 3 & 7
\end{array}\right|\)
Solution:
(i) Let |A| = \(\left|\begin{array}{rrr}
2 & 4 & 1 \\
8 & 5 & 2 \\
-1 & 3 & 7
\end{array}\right|\) ;
expanding along R₁
= \(2\left|\begin{array}{ll}
5 & 2 \\
3 & 7
\end{array}\right|-4\left|\begin{array}{rr}
8 & 2 \\
-1 & 7
\end{array}\right|+1\left|\begin{array}{rr}
8 & 5 \\
-1 & 3
\end{array}\right|\)
= 2 (35 – 6) – 4 (56 + 2) + 1 (24 + 5)
= 58 – 232 + 29
= – 145

Question 12.
Using cofactors of elements of second row, evacuate Δ = \(\left|\begin{array}{lll}
5 & 3 & 8 \\
2 & 0 & 1 \\
1 & 2 & 3
\end{array}\right|\). (NCERT)
Solution:
Given Δ = \(\left|\begin{array}{lll}
5 & 3 & 8 \\
2 & 0 & 1 \\
1 & 2 & 3
\end{array}\right|\)
∴ Δ = \(-2\left|\begin{array}{ll}
3 & 8 \\
2 & 3
\end{array}\right|+0\left|\begin{array}{ll}
5 & 8 \\
1 & 3
\end{array}\right|-1\left|\begin{array}{ll}
5 & 3 \\
1 & 2
\end{array}\right|\)
= – 2 (9 – 16) + 0 (15 – 8) – (10 – 3)
= 14 + 0 – 7
= 7.