Regular engagement with ML Aggarwal Maths for Class 12 Solutions Chapter 4 Determinants Chapter Test can boost students’ confidence in the subject.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Chapter Test

Question 1.
If A and B are square matrices of order 3 such that |A| = – 2 and |B| = 3, find the value of |- 5 AB|.
Solution:
Given A and B are square matrices of order 3 such that
|A| = – 2 and |B| = 3 …………(1)
We know that if A be a square matrix of order n
Then |kA| = kn A …………..(2)
Now |- 5 AB| = (- 5)3 |A| |B| [using (1) and (2)
= – 125 × (- 2) × 3
= 750

Question 2.
Using properties of determinants, prove the following:
(i) $$\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q \end{array}\right|$$ = 1
(ii) $$\left|\begin{array}{lll} p a & p^2+a^2 & 1 \\ p b & p^2+b^2 & 1 \\ p c & p^2+c^2 & 1 \end{array}\right|$$ = p (a – b) (b – c) (c – a)
Solution:
(i) Let Δ = $$\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q \end{array}\right|$$
operate R2 → R2 – 2R1 ;
R3 → R3 – 3R1
= $$\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 0 & 1 & 2+p \\ 0 & 3 & 7+3 p \end{array}\right|$$
Expanding along C1 ; we get
= 1 $$\left|\begin{array}{cc} 1 & 2+p \\ 3 & 7+3 p \end{array}\right|$$
= (7 + 3p) – 3 (2 + p)
= 7 + 3p – 6 – 3p = 1

(ii) Let Δ = $$\left|\begin{array}{lll} p a & p^2+a^2 & 1 \\ p b & p^2+b^2 & 1 \\ p c & p^2+c^2 & 1 \end{array}\right|$$
Taking p common from C1 ; we have
= p $$\left|\begin{array}{ccc} a & p^2+a^2 & 1 \\ b & p^2+b^2 & 1 \\ c & p^2+c^2 & 1 \end{array}\right|$$ ;
operate R2 → R2 – 2R1 ;
R3 → R3 – 3R1
= p $$\left|\begin{array}{ccc} a & p^2+a^2 & 1 \\ b-a & b^2-a^2 & 0 \\ c-a & c^2-a^2 & 0 \end{array}\right|$$
Taking (b – a) common from r2 and (c – a) common from R3
= p (b – a) (c – a) $$\left|\begin{array}{ccc} a & p^2+a^2 & 1 \\ 1 & b+a & 0 \\ 1 & c+a & 0 \end{array}\right|$$ ;
Expanding along C3 ;
= p (b – a) (c – a) (c + a – b – a)
= p (a – b) (b – c) (c – a).

Question 3.
If a, b and c are real numbers and ∆ = $$\left|\begin{array}{lll} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right|$$ = 0, then show that either a + b + c = 0 or a = b = c.
Solution:
Since, we have $$\left|\begin{array}{lll} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right|$$ = 0
operate C1 → C1 + C2 + C3
⇒ $$\left|\begin{array}{lll} 2(a+b+c) & c+a & a+b \\ 2(a+b+c) & a+b & b+c \\ 2(a+b+c) & b+c & c+a \end{array}\right|$$ = 0
Taking 2 (a + b + c) common from C1
2 (a + b + c) $$\left|\begin{array}{lll} 1 & c+a & a+b \\ 1 & a+b & b+c \\ 1 & b+c & c+a \end{array}\right|$$ = 0
operate R2 → R2 – R1 ;
R3 → R3 – R1
⇒ 2 (a + b + c) $$\left|\begin{array}{lll} 1 & c+a & a+b \\ 0 & b-c & c-a \\ 0 & b-a & c-b \end{array}\right|$$ = 0
Expanding along C1 ; we have
⇒ 2 (a + b + c) $$\left|\begin{array}{ll} b-c & c-a \\ b-a & c-b \end{array}\right|$$ = 0
⇒ 2 (a + b + c) [- (b – c)2 – (b – a) (c – a)] = 0
⇒ 2 (a + b + c) [- b2 – c2 + 2bc – bc + ab + ac – a2] = 0
⇒ – 2 (a + b + c) [a2 + b2 + c2 – ab – bc – ca] = 0
⇒ (a + b + c) [2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca] = 0
⇒ (a + b + c) [(a2 + b2 – 2ab) + (b2 + c2 – 2bc) + (c2 + a2 – 2ac)] = 0
⇒ (a + b + c) [(a – b)2 + (b – c)2 + (c – a)2 ]= 0
∴ either a + b + c = 0 or
(a – b)2 + (b – c)2 + (c a)2 = 0
i.e. a + b + c = 0 or
a = b = c.

Question 4.
If p + q + r = 0, prove that $$\left|\begin{array}{lll} p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a \end{array}\right|$$ = pqr $$\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|$$. (NCERT Exampler)
Solution:
L.H.S. = $$\left|\begin{array}{lll} p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a \end{array}\right|$$ ;
expanding along R1
= pa [qra2 – p2bc] – qb [q2ac – prb2] + rc [pqc2 – r2ab]
= pqr (a3 + b3 + c3) – abc (p3 + q3 + r3) ………..(1)
Since p + q + r = 0
⇒ p3 + q3 + r3 = 3pqr
∴ from (1) ; we have
L.H.S. = pqr (a3 + b3 + c3) – 3pqrabc
= pqr (a3 + b3 + c3 – 3abc) ……………(2)
R.H.S. = pqr $$\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|$$ ;
operate C1 → C1 + C2 + C3
= pqr $$\left|\begin{array}{lll} a+b+c & b & c \\ a+b+c & a & b \\ a+b+c & c & a \end{array}\right|$$ ;
Taking (a + b + c) common from C1
= pqr (a + b + c) $$\left|\begin{array}{lll} 1 & b & c \\ 1 & a & b \\ 1 & c & a \end{array}\right|$$ ;
operate R2 → R2 – R1 ;
R3 → R3 – R1
= pqr (a + b + c) $$\left|\begin{array}{rrr} 1 & b & c \\ 0 & a-b & b-c \\ 0 & c-b & a-c \end{array}\right|$$ ;
expanding along C1
= pqr (a + b + c) [(a – b) (a – c) + (b – c)2]
= pqr (a + b + c) [a2 + b2 + c2 – ab – ac – bc]
= pqr [a3 + b3 + c3 – 3abc] …………(3)
From eqn. (2) and eqn. (3) ; we have
L.H.S. = R.H.S.
Thus, $$\left|\begin{array}{lll} p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a \end{array}\right|$$ = pqr $$\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|$$.

Question 5.
(i) Show that $$\left|\begin{array}{ccc} \cos ^2 x & \sin x \cos x & -\sin x \\ \sin x \cos x & \sin ^2 x & \cos x \\ \sin x & -\cos x & 0 \end{array}\right|$$ = 1, for all real values of x.
(ii) Show that $$\left|\begin{array}{ccc} 1 & a & a^2 \\ \cos (n-1) x & \cos n x & \cos (n+1) x \\ \sin (n-1) x & \sin n x & \sin (n+1) x \end{array}\right|$$ is independent of n.
Solution:
Let Δ = $$\left|\begin{array}{ccc} \cos ^2 x & \sin x \cos x & -\sin x \\ \sin x \cos x & \sin ^2 x & \cos x \\ \sin x & -\cos x & 0 \end{array}\right|$$ ;
operate C1 → C1 – sin x C3 ;
C2 → C2 + cos x C3
= $$\left|\begin{array}{ccc} 1 & 0 & -\sin x \\ 0 & 1 & \cos x \\ \sin x & -\cos x & 0 \end{array}\right|$$ ;
expanding along R1
= 1 (0 + cos2 x) + 0 – sin x (0 – sin x)
= cos2 x + sin2 x
= 1 ∀ x ∈ R

(ii) Let Δ = $$\left|\begin{array}{ccc} 1 & a & a^2 \\ \cos (n-1) x & \cos n x & \cos (n+1) x \\ \sin (n-1) x & \sin n x & \sin (n+1) x \end{array}\right|$$ ;
expanding along R1
= 1 [cos nx sin (n + 1) x – sin nx cos (n + 1) x] – a [sin (n + 1) x cos (n – 1) x – sin (n – 1) x cos (n + 1) x] + a2 [sin nx cos (n – 1) x – cos nx sin (n – 1) x]
= sin (n + 1 – n) x – a sin (n + 1 – n + 1) x + a2 sin (n – n + 1) x
[∵ sin (A ± B) = sin A cos B ± cos A sin B)]
= sin x – a sin 2x + a2 sin x
= (1 + a2) sin x – a sin 2x
Thus, Δ is independent of n.

Question 6.
Solve the following for x :
(i) $$\left|\begin{array}{lll} 1 & x & x^3 \\ 1 & a & a^3 \\ 1 & b & b^3 \end{array}\right|$$ = 0, a ≠ b
(ii) $$\left|\begin{array}{ccc} a^2+x & a b & a c \\ a b & b^2+x & b c \\ a c & b c & c^2+x \end{array}\right|$$ = 0.
Solution:
(i) Given, $$\left|\begin{array}{lll} 1 & x & x^3 \\ 1 & a & a^3 \\ 1 & b & b^3 \end{array}\right|$$ = 0
operate R2 → R2 – R1 ;
R3 → R3 – R1
$$\left|\begin{array}{rrr} 1 & x & x^3 \\ 0 & a-x & a^3-x^3 \\ 0 & b-x & b^3-x^3 \end{array}\right|$$ = 0
Taking (a – x) common from R2 and (b – x) common from R3
(a – x) (b – x) $$\left|\begin{array}{ccc} 1 & x & x^3 \\ 0 & 1 & a^2+a x+x^2 \\ 0 & 1 & b^2+b x+x^2 \end{array}\right|$$ = 0 ;
Expanding along C1 ; we have
(a – x) (b – x) $$\left|\begin{array}{ll} 1 & a^2+a x+x^2 \\ 1 & b^2+b x+x^2 \end{array}\right|$$ = 0
⇒ (a – x) (b – x) [b2 + bx + x2 – a2 – ax – a2] = 0
⇒ (a – x) (b – x) [(b – a) (b + a) + x (b – a)]
⇒ (a – x) (b – x) (b – a) (x + b + a) = 0
a – x = 0 or
b – x = 0 or
x + b + a = 0 or
b – a = 0
⇒ x = a or
x = b or
x = – (a + b) [but b ≠ a]
∴ x = a, b, – (a + b)

(ii) Given $$\left|\begin{array}{ccc} a^2+x & a b & a c \\ a b & b^2+x & b c \\ a c & b c & c^2+x \end{array}\right|$$ = 0
Multiply C1 by a, C2 by b and C3 by c ; we get
⇒ $$\frac{1}{a b c}\left|\begin{array}{ccc} a\left(a^2+x\right) & a b^2 & a c^2 \\ a^2 b & b\left(b^2+x\right) & b c^2 \\ a^2 c & b^2 c & c\left(c^2+x\right) \end{array}\right|$$ = 0 ;
operate C1 → C1 + C2 + C3
⇒ $$\frac{1}{a b c}\left|\begin{array}{lcc} a\left(a^2+x+b^2+c^2\right) & a b^2 & a c^2 \\ b\left(a^2+b^2+x+c^2\right) b\left(b^2+x\right) & b c^2 \\ c\left(a^2+b^2+c^2+x\right) & b^2 c & c\left(c^2+x\right) \end{array}\right|$$ = 0
Taking (a2 + x + b2 + c2) common from C1; we have
⇒ $$\frac{\left(a^2+b^2+c^2+x\right)}{a b c}\left|\begin{array}{ccc} a & a b^2 & a c^2 \\ b & b\left(b^2+x\right) & b c^2 \\ c & b^2 c & c\left(c^2+x\right) \end{array}\right|$$ = 0
Taking a, b and c common from R1, R2 and R3 respectively
⇒ (a2 + b2 + c2 + x) $$\left|\begin{array}{ccc} 1 & b^2 & c^2 \\ 1 & b^2+x & c^2 \\ 1 & b^2 & c^2+x \end{array}\right|$$ = 0 ;
operate R2 → R2 – R1 ;
R3 → R3 – R1
⇒ (a2 + b2 + c2 + x) $$\left|\begin{array}{ccc} 1 & b^2 & c^2 \\ 0 & x & 0 \\ 0 & 0 & x \end{array}\right|$$ = 0 ;
Expanding along C1 ; we have
⇒ (x + a2 + b2 + c2) x2 = 0
⇒ x = 0, 0, – (a2 + b2 + c2)

Question 7.
(i) Factorise $$\left|\begin{array}{ccc} x^2 & (y-z)^2 & y z \\ y^2 & (z-x)^2 & z x \\ z^2 & (x-y)^2 & x y \end{array}\right|$$.
(ii) $$\left|\begin{array}{ccc} 0 & x y z & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0 \end{array}\right|$$ (Exampler)
Solution:

Taking (x – y), (x – z) common from R2 and R3 respectively
= (x2 + y2 + z2) (x – y) (x – z) $$\left|\begin{array}{ccc} 1 & y^2+z^2 & y z \\ 0 & x+y & z \\ 0 & x+z & y \end{array}\right|$$
= (x2 + y2 + z2) (x – y) (x – z) [xy + y2 – zx – z2]
= (x2 + y2 + z2) (x – y) (x – z) [x (y – z) + (y – z) (y + z)]
= (x2 + y2 + z2) (x – y) (x – z) (y – z) (x + y + z)
= – (x2 + y2 + z2) (x – y) (y – z) (z – x) (x + y + z)

(ii) Let ∆ = $$\left|\begin{array}{ccc} 0 & x y z & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0 \end{array}\right|$$ ;
operate R2 → R2 – R3
= $$\left|\begin{array}{ccc} 0 & x y z & x-z \\ y-z & y-z & y-z \\ z-x & z-y & 0 \end{array}\right|$$ ;
Taking (y – z) common from R2
= (y – z) $$\left|\begin{array}{ccc} 0 & x y z & x-z \\ 1 & 1 & 1 \\ z-x & z-y & 0 \end{array}\right|$$ ;
operate C2 → C2 – C1
C3 → C3 – C1
= (y – z) $$\left|\begin{array}{ccc} 0 & x y z & x-z \\ 1 & 0 & 0 \\ z-x & x-y & x-z \end{array}\right|$$
= (y – z) (z – x) $$\left|\begin{array}{ccc} 0 & x y z & 1 \\ 1 & 0 & 0 \\ z-x & x-y & 1 \end{array}\right|$$
Expanding along R2
= – (y – z) (x – z) (xyz – x + y)
= (y – z) (z – x) (xyz + y – z)

Question 7 (old).
Prove that $$\left|\begin{array}{ccc} a & b & c \\ a\left(a^2+1\right) & b\left(b^2+1\right) & c\left(c^2+1\right) \\ a+1 & b+1 & c+1 \end{array}\right|$$ = (a – b) (b – c) (c – a) (a + b + c)
Solution:
Let = $$\left|\begin{array}{ccc} a & b & c \\ a\left(a^2+1\right) & b\left(b^2+1\right) & c\left(c^2+1\right) \\ a+1 & b+1 & c+1 \end{array}\right|$$ ;
operate R2 → R2 – R1
R3 → R3 – R1
= $$\left|\begin{array}{ccc} a & b & c \\ a^3 & b^3 & c^3 \\ 1 & 1 & 1 \end{array}\right|$$ ;
operate C2 → C2 – C1 ;
C3 → C3 – C1
= $$\left|\begin{array}{ccc} a & b-a & c-a \\ a^3 & b^3-a^3 & c^3-a^3 \\ 1 & 0 & 0 \end{array}\right|$$
Taking (b – a) common from C2 and (c – a) common from C3 respectively
= (b – a) (c – a) $$\left|\begin{array}{ccc} a & 1 & 1 \\ a^3 & b^2+a b+a^2 & c^2+a c+a^2 \\ 1 & 0 & 0 \end{array}\right|$$ ;
expanding along R3
= (b – a) (c – a) [c2 + ac + a2 – b2 – ab – a2]
= (b – a) (c – a) [(c2 – b2) + a (c – b)]
= (b – a) (c – a) [(c – b) (c + b) + a (c – b)]
= (b – a) (c – a) (c – b) (c + b + a)
= (a – b) (b – c) (c – a) (a + b + c)

Question 8.
If Δ = $$\left|\begin{array}{ccc} (x-2)^2 & (x-1)^2 & x^2 \\ (x-1)^2 & x^2 & (x+1)^2 \\ x^2 & (x+1)^2 & (x+2)^2 \end{array}\right|$$, then show that Δ is a negative even integer.
Solution:
Given Δ = $$\left|\begin{array}{ccc} (x-2)^2 & (x-1)^2 & x^2 \\ (x-1)^2 & x^2 & (x+1)^2 \\ x^2 & (x+1)^2 & (x+2)^2 \end{array}\right|$$
operate C1 → C1 – C3 ;
C2 → C2 – C3
= $$\left|\begin{array}{ccc} -4 x+4 & -2 x+1 & x^2 \\ -4 x & -2 x-1 & (x+1)^2 \\ -4 x-4 & -2 x-3 & (x+2)^2 \end{array}\right|$$
Taking – 4 common from C1
= – 4 $$\left|\begin{array}{ccc} x-1 & -2 x+1 & x^2 \\ x & -2 x-1 & (x+1)^2 \\ x+1 & -2 x-3 & (x+2)^2 \end{array}\right|$$ ;
operate R2 → R2 – R1
R3 → R3 – R1
= – 4 $$\left|\begin{array}{ccc} x-1 & -2 x+1 & x^2 \\ 1 & -2 & 2 x+1 \\ 2 & -4 & 4 x+4 \end{array}\right|$$ ;
operate R3 → R3 – 2R2
= – 4 $$\left|\begin{array}{ccc} x-1 & -2 x+1 & x^2 \\ 1 & -2 & 2 x+1 \\ 0 & 0 & 2 \end{array}\right|$$ ;
expanding along R3
∴ Δ = – 4 × 2 $$\left|\begin{array}{cc} x-1 & -2 x+1 \\ 1 & -2 \end{array}\right|$$
= – 8 [- 2x + 2 + 2x – 1]
= – 8
Hence Δ be clearly a negative even integer.

Question 9.
If x, y, z are all different and $$\left|\begin{array}{ccc} x^3 & (x+p)^3 & (x-p)^3 \\ y^3 & (y+p)^3 & (y-p)^3 \\ z^3 & (z+p)^3 & (z-p)^3 \end{array}\right|$$ = 0, prove that p2 (x + y + z) = 3xyz.
Solution:

Taking (y – x), (z – x) common from R2 and R3 resp. and expanding along C3 ; we have
= xyz (y – x) (z – x) $$\left|\begin{array}{ll} y+x & 1 \\ z+x & 1 \end{array}\right|$$
= xyz (y – x) (y – z) (z – x) ………….(3)
putting (2) and (3) in eqn. (1) ; we get
∴ 0 = 6p5 (x – y) (y – z) (z – x) (x + y + z) – 183 xyz (x – y) (y – z) (z – x)
[∴ Δ = 0 given]
i.e. 0 = (x – y) (y – z) (z – x) [p2 (x + y + z) – 3xyz]
since x, y, z are all different i.e. x ≠ y ≠ z
∴ (x – y) (y – z) (z – x) ≠ 0
i.e. p2 (x + y + z) = 3xyz is the required result.

Question 10.
If a2 + b2 + c2 = – 2 and f(x) = $$\left|\begin{array}{ccc} 1+a^2 x & \left(1+b^2\right) x & \left(1+c^2\right) x \\ \left(1+a^2 x\right) & 1+b^2 x & \left(1+c^2\right) x \\ \left(1+a^2 x\right) & \left(1+b^2\right) x & 1+c^2 x \end{array}\right|$$, then prove that f(x) is a polynomial of degree 2.
Solution:
Given f(x) = $$\left|\begin{array}{ccc} 1+a^2 x & \left(1+b^2\right) x & \left(1+c^2\right) x \\ \left(1+a^2 x\right) & 1+b^2 x & \left(1+c^2\right) x \\ \left(1+a^2 x\right) & \left(1+b^2\right) x & 1+c^2 x \end{array}\right|$$ ;
operate C1 → C1 + C2 + c3
= $$\left|\begin{array}{ccc} 1+2 x+\left(a^2+b^2+c^2\right) x & \left(1+b^2\right) x & \left(1+c^2\right) x \\ 1+2 x+\left(a^2+b^2+c^2\right) x & 1+b^2 x & \left(1+c^2\right) x \\ 1+2 x+\left(a^2+b^2+c^2\right) x & \left(1+b^2\right) x & 1+c^2 x \end{array}\right|$$
since a2 + b2 + c2 = – 2 (given)
= $$\left|\begin{array}{ccc} 1 & \left(1+b^2\right) x & \left(1+c^2\right) x \\ 1 & 1+b^2 x & \left(1+c^2\right) x \\ 1 & \left(1+b^2\right) x & 1+c^2 x \end{array}\right|$$
operate R2 → R2 – R1 ;
R3 → R3 – R1
= $$\left|\begin{array}{ccc} 1 & \left(1+b^2\right) x & \left(1+c^2\right) x \\ 0 & 1-x & 0 \\ 0 & 0 & 1-x \end{array}\right|$$ ;
Expanding along C1
= (1 – x)2 which is clearly a polynomial of degree 2.

Question 11.
If a, b, c are all different and $$\left|\begin{array}{ccc} 1 & 1 & 1 \\ (x-a)^2 & (x-b)^2 & (x-c)^2 \\ (x-b)(x-c) & (x-c)(x-a) & (x-a)(x-b) \end{array}\right|$$ = 0, then find the value(s) of x.
Solution:
Let Δ = $$\left|\begin{array}{ccc} 1 & 1 & 1 \\ (x-a)^2 & (x-b)^2 & (x-c)^2 \\ (x-b)(x-c) & (x-c)(x-a) & (x-a)(x-b) \end{array}\right|$$ ………………(1)
putting x – a = α,
x – b = β and
x – c = γ in eqn. (1) ; we get
= $$\left|\begin{array}{ccc} 1 & 1 & 1 \\ \alpha^2 & \beta^2 & \gamma^2 \\ \beta \gamma & \gamma \alpha & \alpha \beta \end{array}\right|$$ ;
Operate C2 → C2 – C1 ;
C3 → C3 – C1
= $$\left|\begin{array}{ccc} 1 & 0 & 0 \\ \alpha^2 & \beta^2-\alpha^2 & \gamma^2-\alpha^2 \\ \beta \gamma & \gamma(\alpha-\beta) & \beta(\alpha-\gamma) \end{array}\right|$$
Taking (β – α) common from C2 and (γ – α) common from C3
= (β – α) (γ – α) $$\left|\begin{array}{ccc} 1 & 0 & 0 \\ \alpha^2 & \beta+\alpha & \gamma+\alpha \\ \beta \gamma & -\gamma & -\beta \end{array}\right|$$ ;
Expanding along R1 ; we have
= (β – α) (γ – α) $$\left|\begin{array}{cc} \beta+\alpha & \gamma+\alpha \\ -\gamma & -\beta \end{array}\right|$$
= (β – α) (γ – α) [- β (β + α) + γ (γ + α)]
= (β – α) (γ – α) [γ2 – β2 + α (γ – β)]
= (β – α) (γ – α) (γ – β) (α + β + γ)
= (α – β) (β – γ) (γ – α) (α + β + γ)
Also Δ has value given to be 0.
∴ (α – β) (β – γ) (γ – α) (α + β + γ) = 0
⇒ [(x – a) – (x – b) [(x – b) – (x – c)] [(x – c) – (x – a)] [3x – a – b – c] = 0
⇒ (b – a) (c – b) (a – c) (3x – a – b – c) = 0
⇒ a ≠ b ≠ c
⇒ a – b ≠ 0, b – c ≠ 0, c – a ≠ 0
⇒ (b – a) (c – b) (c – a) ≠ 0
∴ from (2) ; 3x – a – b – c = 0
⇒ x = $$\frac{a+b+c}{3}$$

Question 12.
If A = $$\left[\begin{array}{cc} 1 & \tan \frac{x}{2} \\ -\tan \frac{x}{2} & 1 \end{array}\right]$$, then show that A’A-1 = $$\left[\begin{array}{rr} \cos x & -\sin x \\ \sin x & \cos x \end{array}\right]$$.
Solution:
Given A = $$\left[\begin{array}{cc} 1 & \tan \frac{x}{2} \\ -\tan \frac{x}{2} & 1 \end{array}\right]$$
∴ |A| = 1 + tan2 $$\frac{x}{2}$$
= sec2 $$\frac{x}{2}$$ ≠ 0 ∀ x ≠ odd multiple of π
Here A11 = 1 ;
A12 = tan $$\frac{x}{2}$$
A21 = – tan $$\frac{x}{2}$$
A22 = 1
∴ adj A = $$\left[\begin{array}{cc} 1 & \tan \frac{x}{2} \\ -\tan \frac{x}{2} & 1 \end{array}\right]$$
= $$\left[\begin{array}{cc} 1 & -\tan \frac{x}{2} \\ \tan \frac{x}{2} & 1 \end{array}\right]$$
Since |A| ≠ 0
∴ A-1 exists
and A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A
∴ A-1 = $$\frac{1}{1+\tan ^2 \frac{x}{2}}\left[\begin{array}{cc} 1 & -\tan \frac{x}{2} \\ \tan \frac{x}{2} & 1 \end{array}\right]$$
Thus A’A-1 = $$\frac{1}{1+\tan ^2 \frac{x}{2}}\left[\begin{array}{cc} 1 & -\tan \frac{x}{2} \\ \tan \frac{x}{2} & 1 \end{array}\right]\left[\begin{array}{cc} 1 & -\tan \frac{x}{2} \\ \tan \frac{x}{2} & 1 \end{array}\right]$$

= $$=\frac{1}{1+\tan ^2 \frac{x}{2}}\left[\begin{array}{cc} 1-\tan ^2 \frac{x}{2} & -2 \tan \frac{x}{2} \\ 2 \tan \frac{x}{2} & 1-\tan ^2 \frac{x}{2} \end{array}\right]$$

= $$\left[\begin{array}{cc} \frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}} & \frac{-2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}} \\ \frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}} & \frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}} \end{array}\right]$$

= $$\left[\begin{array}{rr} \cos x & -\sin x \\ \sin x & \cos x \end{array}\right]$$

Question 13.
For the matrix A = $$\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right]$$, show that A2 – 4A + 5I = O. Hence, obtain A-1.
Solution:
Given A = $$\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right]$$
Here |A| = 3 + 2 = 5 ≠ 0
∴ A-1 exists
L.H.S. = A2 – 4A + 5I
= $$\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right]\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right]-4\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]+5\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right]-\left[\begin{array}{cc} 4 & -4 \\ 8 & 12 \end{array}\right]+\left[\begin{array}{cc} 5 & 0 \\ 0 & 5 \end{array}\right]$$
= $$\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$$
= O = R.H.S.
Thus A2 – 4A + 5I = O
⇒ 5I = 4A – A2 …………….(1)
pre-multiplying both sides of eqn. (1) by A-1 ; we have
5A-1I = 4A-1 A – (A-1 A) A
⇒ 5A-1 = 4I – IA
[∵ A-1 A = I]
⇒ 5A-1 = $$4\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right]$$
⇒ 5A-1 = $$\left[\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}\right]-\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]$$
= $$\left[\begin{array}{rr} 3 & 1 \\ -2 & 1 \end{array}\right]$$
⇒ A-1 = $$\frac{1}{5}\left[\begin{array}{rr} 3 & 1 \\ -2 & 1 \end{array}\right]$$

Question 14.
Using matrix method, solve the following system of linear equations:
(i) 3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4

(ii) 2x + y + z = 1
x – 2y – z = $$\frac{3}{2}$$
3y – 5z = 9.
Solution:
(i) Given system of equations is equivalent to AX = B
where A = $$\left[\begin{array}{rrr} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{array}\right]$$ ;
X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ ;
B = $$\left[\begin{array}{l} 8 \\ 1 \\ 4 \end{array}\right]$$
Here |A|= $$\left|\begin{array}{rrr} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{array}\right|$$
= 3 (2 – 3) + 2 (4 + 4) + 3 (- 6 – 4)
= – 3 + 16 – 30
= – 17 ≠ 0
∴ A-1 exists and given system has unique solution.
The cofactors of R1 are

Thus x = 1 ; y = 2 and z = 3
Hence, the required solution of given system be x = 1, y = 2 and z = 3.

(ii) Given system of equations is equivalent to AX = B
where A = $$\left[\begin{array}{rrr} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{array}\right]$$ ;
X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$
and B = $$\left[\begin{array}{r} 1 \\ 3 / 2 \\ 9 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{rrr} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{array}\right|$$ ;
Expanding along R1
= 2 (10 + 3) – 1 (- 5 – 0) + 1 (3 – 0)
= 26 + 5 + 3
= 34 ≠ 0
∴ A-1 exists and given system be unique solution.

Question 15.
If A = $$\left[\begin{array}{rrr} 3 & -2 & 1 \\ 2 & 1 & -3 \\ -1 & 2 & 1 \end{array}\right]$$, find A-1. Using A, solve the following system of equations:
3x – 2y + z = 2, 2x + y – 3z = – 5, – x + 2y + z = 6.
Solution:
Given A = $$\left[\begin{array}{rrr} 3 & -2 & 1 \\ 2 & 1 & -3 \\ -1 & 2 & 1 \end{array}\right]$$
Here |A| = $$\left|\begin{array}{rrr} 3 & -2 & 1 \\ 2 & 1 & -3 \\ -1 & 2 & 1 \end{array}\right|$$
= 3 (1 + 6) + 2 (2 – 3) + 1 (4 + 1)
= 21 – 2 + 5
= 24 ≠ 0
∴ A-1 exists and
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 A = $$\left[\begin{array}{rrr} 3 & -2 & 1 \\ 2 & 1 & -3 \\ -1 & 2 & 1 \end{array}\right]$$
X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$
and B = $$=\left[\begin{array}{r} 2 \\ -5 \\ 6 \end{array}\right]$$
Here |A| = 24 ≠ 0
∴ given system has unique solution given by
X = A-1B
⇒ X = $$\frac{1}{24}\left[\begin{array}{rrr} 7 & 4 & 5 \\ 1 & 4 & 11 \\ 5 & -4 & 7 \end{array}\right]\left[\begin{array}{r} 2 \\ -5 \\ 6 \end{array}\right]$$
⇒ $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{24}\left[\begin{array}{r} 14-20+30 \\ 2-20+66 \\ 10+20+42 \end{array}\right]$$
⇒ $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{24}\left[\begin{array}{l} 24 \\ 48 \\ 72 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]$$
Thus x = 1 ; y = 2 and z = 3.

Question 16.
Evaluate $$\left[\begin{array}{rrr} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{array}\right]\left[\begin{array}{rrr} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{array}\right]$$. Hence, solve the system of equations
3x – 2y + 3z = 8, 2x + y = 1, 4x – 3y + 2z = 4
Solution:
Let A = $$\left[\begin{array}{rrr} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{array}\right]$$
and B = $$\left[\begin{array}{rrr} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{array}\right]$$
Here AB = $$\left[\begin{array}{rrr} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{array}\right]\left[\begin{array}{rrr} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{array}\right]$$
= $$\left[\begin{array}{rrr} -3+16-30 & -15+12+3 & -3-18+21 \\ -2-8+10 & -10-6-1 & -2+9-7 \\ -4+24-20 & -20+18+2 & -4-27+14 \end{array}\right]$$

Expanding along R1
= 3 (2 – 3) + 2 (4 + 4) + 3 (- 6 – 4)
= – 3 + 16 – 30
= – 17 ≠ 0
∴ A-1 exists.
Now AX = C
⇒ X = A-1C
⇒ $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=-\frac{1}{17}\left[\begin{array}{rrr} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{array}\right]\left[\begin{array}{l} 8 \\ 1 \\ 4 \end{array}\right]$$
= $$-\frac{1}{17}\left[\begin{array}{r} -8-5-4 \\ -64-6+36 \\ -80+1+28 \end{array}\right]$$
⇒ $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=-\frac{1}{17}\left[\begin{array}{l} -17 \\ -34 \\ -51 \end{array}\right]$$
= $$\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]$$
∴ x = 1 ; y = 2 and z = 3.

Question 17.
Examine whether or not the system of equations 2x – 5y = 3, 4x – 10y = 6 is inconsistent. If consistent, solve it.
Solution:
Given system of eqns is 2x – 5y = 3, 4x – 10y = 6
Here D = $$\left|\begin{array}{rr} 2 & -5 \\ 4 & -10 \end{array}\right|$$
= – 20 + 20 = 0
D1 = $$\left|\begin{array}{cc} 3 & -5 \\ 6 & -10 \end{array}\right|$$
= – 30 + 30 = 0
D2 = $$\left|\begin{array}{ll} 2 & 3 \\ 4 & 6 \end{array}\right|$$
= 12 – 12 = 0
Thus, D = 0 = D1 = D2
∴ Given system of eqn’s has infinitely many solutions and given system is consistent.

For solution:
putting y = k in first eqn. of given system , we get
2x – 5k = 3
⇒ x = $$\frac{3+5 k}{2}$$
Clearly x = $$\frac{3+5 k}{2}$$, y = k satisfies the second eqn. of given system.
Hence the required solution of given system be x = $$\frac{3+5 k}{2}$$; y = k , where k be any number.

Question 18.
Examine whether or not the system of equations
x – y + z = 3
2x + y – z = 2
x + 2y – 2z = – 1
Here, D = $$\left|\begin{array}{rrr} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & 2 & -2 \end{array}\right|$$
= 1 (- 2 + 2) + 1 (- 4 + 1) + 1 (4 – 1)
= 0 – 3 + 3 = 0
D1 = $$\left|\begin{array}{rrr} 3 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & 2 & -2 \end{array}\right|$$
= 3 (- 2 + 2) + 1 (- 4 – 1) + 1 (4 + 1)
= 0 – 5 + 5 = 0
D2 = $$\left|\begin{array}{rrr} 1 & 3 & 1 \\ 2 & 2 & -1 \\ 1 & -1 & -2 \end{array}\right|$$
= 1 (- 4 – 1) – 3 (- 4 + 1) + 1 (- 2 – 2)
= – 5 + 9 – 4 = 0
and D3 = $$\left|\begin{array}{rrr} 1 & -1 & 3 \\ 2 & 1 & 2 \\ 1 & 2 & -1 \end{array}\right|$$
= 1 (- 1 – 4) – 3 (- 4 + 1) + 3 (4 – 1)
= – 5 – 4 + 9 = 0
Thus D = D3 = D1 = D2 = 0
Then by cramer’s rule, given system is consistent and have infinitely many solutions.

For solution:
putting z = k, where k be any real number in first two eqn’s of given system, we get
x – y = 3 – k
2x + y = 2 + k
Here |A| = $$\left|\begin{array}{rr} 1 & -1 \\ 2 & 1 \end{array}\right|$$
= 1 + 2
= 3 ≠ 0
∴ given system has unique soln.
|A1| = $$\left|\begin{array}{rr} 3-k & -1 \\ 2+k & 1 \end{array}\right|$$
= 3 – k + 2 + k = 5
|A2| = $$\left|\begin{array}{cc} 1 & 3-k \\ 2 & 2+k \end{array}\right|$$
= 2 + k – 6 + 2k
= 3k – 4
Then by Cramer’s rule, we have
x = $$\frac{\left|A_1\right|}{|A|}=\frac{5}{3}$$
and y = $$\frac{\left|\mathrm{A}_2\right|}{|\mathrm{A}|}=\frac{3 k-4}{3}$$
= k – $$\frac{4}{3}$$
Also x = $$\frac{5}{3}$$, y = k – $$\frac{4}{3}$$ and z = k,
where k be any real number satisfies the 3rd eqn. of given system.
Hence required soln. of given system be x = $$\frac{5}{3}$$, y = k – $$\frac{4}{3}$$ and z = k,
where k be any real number.