PSEB 12th Class Maths Solutions Chapter 4 Determinants Ex 4.2

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 4 Determinants Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2

Direction (1 – 5): Using the property of determinants and without expanding.

Question 1.
$\left|\begin{array}{lll} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{array}\right|$ = 0
Solution.
We have, $\left|\begin{array}{lll} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{array}\right|=\left|\begin{array}{lll} x & a & a \\ y & b & b \\ z & c & c \end{array}\right|$ = 0 [applying C₃ → C₃ – C₁]

[Since, the two columns of the determinants are identical.]

Question 2.
$\left|\begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array}\right|$ = 0
Solution.
∆ = $\left|\begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array}\right|$

[applying R₁ → R + R₂]

∆ = $\left|\begin{array}{ccc} a-c & b-a & c-b \\ b-c & c-a & a-b \\ -(a-c) & -(b-a) & -(c-b) \end{array}\right|=-\begin{array}{ccc} a-c & b-a & c-b \\ b-c & c-a & a-b \\ a-c & b-a & c-b \end{array} \mid$ = 0
[Since the two rows R₁ and R₃ are identical].

Question 3.
$\left|\begin{array}{lll} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{array}\right|$ = 0
Solution.
$\left|\begin{array}{lll} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{array}\right|=\left|\begin{array}{lll} 2 & 7 & 2 \\ 3 & 8 & 3 \\ 5 & 9 & 5 \end{array}\right|$ = 0 (applying C₃ → C₃ – 9 C₂)
[Since, two columns are identical]

Question 4.
$\left|\begin{array}{lll} 1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b) \end{array}\right|$ = 0
Solution.
∆ = $\left|\begin{array}{lll} 1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b) \end{array}\right|$

∆ = $\left|\begin{array}{lll} 1 & b c & a b+b c+c a \\ 1 & c a & a b+b c+c a \\ 1 & a b & a b+b c+c a \end{array}\right|$ (applying C₃ → C₃ – 9 C₂)

On taking (ab + bc + ca) common from C₃, we have

∆ = (ab + bc + ca) $\left|\begin{array}{ccc} 1 & b c & 1 \\ 1 & c a & 1 \\ 1 & a b & 1 \end{array}\right|$
= 0 × (ab + bc + ca) = 0
[Since, two columns C₁ and C₃ are identical]

Question 5.
$\left|\begin{array}{lll} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right|=2\left|\begin{array}{lll} a & p & x \\ b & q & y \\ c & r & z \end{array}\right|$
solution.

Direction (6 – 14): By using properties of determinants.

Question 6.
$\left|\begin{array}{ccc} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{array}\right|$ = 0
Solution.
We have, ∆ = $\left|\begin{array}{ccc} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{array}\right|$

∆ = $=\frac{1}{c}\left|\begin{array}{ccc} 0 & a c & -b c \\ -a & 0 & -c \\ b & c & 0 \end{array}\right|$ (applying R₁ → cR₁)

∆ = $\frac{1}{c}\left|\begin{array}{ccc} a b & a c & 0 \\ -a & 0 & -c \\ b & c & 0 \end{array}\right|=\frac{a}{c}\left|\begin{array}{ccc} b & c & 0 \\ -a & 0 & -c \\ b & c & 0 \end{array}\right|$ = 0
(applying R₁ → R₁ – bR₂)
[Since, the two rows R₁ and R₃ are identical].

Question 7.
$\left|\begin{array}{ccc} -a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2} \end{array}\right|$ = 4 a² b² c²
Solution.

[Applying R₂ → R₂ + R₁ and R₃ R₃ + R₁], we have

∆ = a² b² c² $\left|\begin{array}{ccc} -1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 2 & 0 \end{array}\right|$

= a² b² c² $\left|\begin{array}{ll} 0 & 2 \\ 2 & 0 \end{array}\right|$
= a² b² c² (0 – 4)
= 4 a² b² c² = R.H.S.

Question 8.
(i) $\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|$ = (a – b) (b – c) (c – a)

(ii) $\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array}\right|$ = (a – b) (b – c) (c – a) (a + b + c)
Solution.
(i)

By expanding along C₁, we have

∆ = (a – b) (b – c) (c – a) $\left|\begin{array}{cc} 0 & -1 \\ 1 & b+c \end{array}\right|$
= (a – b) (b – c) (c – a)

(ii) Let ∆ = $\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array}\right|$
Applying C₁ → C₁ – C₃, C₂ → C₂ – C₃, we have

By expanding along C₁, we have
∆ = (a – b) (b – c) (c – a) (a + b + c) (- 1) $\left|\begin{array}{cc} 0 & 1 \\ 1 & c \end{array}\right|$
= (a – b) (b – c) (c – a) (a + b + c)

Question 9.
$\left|\begin{array}{lll} x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y \end{array}\right|$ = (x – y) (y – z) (z – x) (xy + yz + zx)
Solution.
L.H.S = $\mid \begin{array}{lll} x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y \end{array}$

Applying R₂ → R₂ – R₁ and R₃ → R₃ – R₁, we have

∆ = $\left|\begin{array}{ccc} x & x^{2} & y z \\ y-x & y^{2}-x^{2} & z x-y z \\ z-x & z^{2}-x^{2} & x y-y z \end{array}\right|$

= (x – y) (z – x) (z – y) [(- xz – yz) + (- x² – xy + x²)]
= – (x – y) (z – x) (z – y) (xy + yz + zx)
= (x – y) (y – z) (z – x) (xy + yz + zx) = R.H.S.
Hence, the given result is proved.

Question 10.
(i) $\left|\begin{array}{ccc} x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4 \end{array}\right|$ = (5x + 4) (4 – x)²

(ii) $\left|\begin{array}{ccc} y+k & y & y \\ y & y+k & y \\ y & y & y+k \end{array}\right|$ = k² (3y + k)
Solution.
(i)

= (5x + 4)(4 – x) (4 – x) $\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 x & 1 & 0 \\ 2 x & 0 & 1 \end{array}\right|$
By expanding along C₃, have
∆ = (5x + 4) (4 – x)² $\left|\begin{array}{cc} 1 & 0 \\ 2 x & 1 \end{array}\right|$
= (5x + 4) (4 – x)²
Hence, the given result is proved.

(ii)

Question 11.
(i) $\left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|$ = (a + b + c)²

(ii) $\left|\begin{array}{ccc} x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y \end{array}\right|$ = 2(x + y + z)²
Solution.
(i) Let ∆ = $\left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|$

Applying R₁ → R₁ + R₂ + R₃, we have
∆ = $\left|\begin{array}{ccc} a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|$

= (a + b + c) $\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|$ (Taking out (a + b+ c) common from R₁)

Applying C₂ → C₂ – C₁ and C₃ → C₃ – C₁, we have
∆ = (a + b + c) $\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 b & -(a+b+c) & 0 \\ 2 c & 0 & -(a+b+c) \end{array}\right|$

= (a + b + c)³ $\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 b & -1 & 0 \\ 2 c & 0 & -1 \end{array}\right|$

By expanding along C₃, we have
∆ = (a + b + c)³ (- 1) (- 1) = (a + b + c)³
Hence, the given result is proved.

(ii)

By expanding along R₃, we have
= 2(x + y+ z)² (1)(1 – 0) = 2(x + y+ z)³
Hence, the given result is proved.

Question 12.
$\left|\begin{array}{ccc} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|$ = (1 – x³)²
Solution.
∆ = $\left|\begin{array}{ccc} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|$

Applying R₁ → R₁ + R₂ + R₃, we have

Question 13.
$\left|\begin{array}{ccc} 1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2} \end{array}\right|$ = (1 + a² + b²)³
Solution.
Let ∆ = $\left|\begin{array}{ccc} 1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2} \end{array}\right|$

Applying R₁ → R₁ + bR₂ and R₂ → R₂ – aR₃

∆ = $\left|\begin{array}{ccc} 1+a^{2}+b^{2} & 0 & -b\left(1+a^{2}+b^{2}\right) \\ 0 & 1+a^{2}+b^{2} & a\left(1+a^{2}+b^{2}\right) \\ 2 b & -2 a & 1-a^{2}-b^{2} \end{array}\right|$

= (1 + a² + b²)² $\left|\begin{array}{ccc} 1 & 0 & -b \\ 0 & 1 & a \\ 2 b & -2 a & 1-a^{2}-b^{2} \end{array}\right|$

By expanding along R₁, we have
∆ = (1 + a² + b²)² $\left[(1)\left|\begin{array}{cc} 1 & a \\ -2 a & 1-a^{2}-b^{2} \end{array}\right|-b\left|\begin{array}{cc} 0 & 1 \\ 2 b & -2 a \end{array}\right|\right]$

= (1 + a² + b²)² [1 – a² – b² + 2a² – b(- 2b)]
= (1 + a² + b²)² (1 + a² + b²)
= (1 + a² + b²)³

Hence, the given result is proved.

Question 14.
$\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|$ = 1 + a² + b² + c²
Solution.

By expanding along R₃, we have
∆ = $-1\left|\begin{array}{cc} b^{2} & c^{2} \\ 1 & 0 \end{array}\right|+1\left|\begin{array}{cc} a^{2}+1 & b^{2} \mid \\ -1 & 1 \end{array}\right|$

= – 1(- c²) + (a² + 1 + b²)
= 1 + a² + b² + c²

Direction (15 – 16): Choose the correct answer in the following questions.

Question 15.
Let A be a square matrix of order 3 × 3 then |KA| is equal to
(A) k |A|
(B) k² |A|
(C) k³ |A|
(D) 3k |A|
Solution.
A is a square matrix of order 3 × 3. (Given)

Question 16.
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) None of these
Solution.
We know that to every square matrix, A [a_ij] of order n. We can associate a number called the determinant of square matrix A, where a_ij = (i, j)th element of A.
Thus, the determinant is a number associated to a square matrix. Hence, the correct answer is (C).