GSEB Solutions Class 12 Maths Chapter 4 Determinants Ex 4.2

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 4 Determinants Ex 4.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 4 Determinants Ex 4.2

Question 1.
$\left|\begin{array}{lll} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{array}\right|$ = 0
Solution:
Operating C₁ → C₁ + C₂, we get
$\left|\begin{array}{lll} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{array}\right|$ = $\left|\begin{array}{lll} x+a & a & x+a \\ y+b & b & y+b \\ z+c & c & z+c \end{array}\right|$ = 0,
since 1st and 3rd columns 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:
Operating C₁ → C₁ + C₂ + C₃ we get
$\left|\begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array}\right|$ = $\left|\begin{array}{ccc} a-b+b-c+c-a & b-c & c-a \\ b-c+c-a+a-b & c-a & a-b \\ c-a+a-b+b-c & a-b & b-c \end{array}\right|$
= $\left|\begin{array}{lll} 0 & b-c & c-a \\ 0 & c-a & a-b \\ 0 & a-b & b-c \end{array}\right|$ = 0,
since all elements of first column are zero.

Question 3.
$\left|\begin{array}{lll} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{array}\right|$ = 0
Solution:
Operating C₃ → C₃ – C₁ – 9C₂ we get:
$\left|\begin{array}{lll} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{array}\right|$ = $\left|\begin{array}{lll} 2 & 7 & 0 \\ 3 & 8 & 0 \\ 5 & 9 & 0 \end{array}\right|$ = 0
since all the elements of 3rd column are zero.

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:
Operating C₃ → C₃ + C₂, we get:
$\left|\begin{array}{lll} 1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b) \end{array}\right|$ = $\left|\begin{array}{lll} 1 & b c & a b+b c+c a \\ 1 & c a & a b+b c+c a \\ 1 & a b & a b+b c+c a \end{array}\right|$
Taking out ab + bc + ca common from 3rd column, we get
= (ab + bc + ca)$\left|\begin{array}{lll} 1 & b c & 1 \\ 1 & c a & 1 \\ 1 & a b & 1 \end{array}\right|$ = 0.
since 1st and 3rd columns are identical.

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

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

Question 7.
$\left|\begin{array}{ccc} -a^{2} & a b & a c \\ b a & -b^{2} & b c \\ a c & c b & -c^{2} \end{array}\right|$ = 4a²b²c².
Solution:
Taking a, b, c common from I, II and III rows respectively,
we get ∆ = $\left|\begin{array}{ccc} -a & b & c \\ a & -b & c \\ a & b & -c \end{array}\right|$
Again a, b and c are taking out common from I, II and III columns respectively. we get
∴ ∆ = a²b²c²$\left|\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right|$
Now operating R₁ → R₁ + R₂, we get:
∆ = a²b²c²$\left|\begin{array}{ccc} 0 & 0 & 2 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right|$ = a²b²c².2(1+1) = 4a²b²c².

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)

Taking out a – b and b – c common from R₁ and R₂ respectively,
∆ = (a-b)(b-c)$\left|\begin{array}{ccc} 0 & 1 & a+b \\ 0 & 1 & b+c \\ 1 & c & c^{2} \end{array}\right|$
Expanding with the help of elements of first common, we get
∆ = (a-b)(b-c)$\left|\begin{array}{ll} 1 & a+b \\ 1 & b+c \end{array}\right|$
⇒ ∆ = (a – b)(b – c)[(b + c) – (a + b)]
⇒ ∆ = (a – b)(b – c)(c – a)

(ii)

Taking out a – b and b – c common from C₁ and C₂ respectively,
∆ = (a-b)(b-c)$\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 1 & c \\ a^{2}+a b+b^{2} & b^{2}+b c+c^{2} & c^{3} \end{array}\right|$
= (a-b)(b-c) x [(b₂ + bc + c²) – (a² + ab + b²)]
= (a – b)(b – c)[(bc – ab) + (c² – a²)]
= (a- b)(b – c)[b(c – a) + (c – a)(c + a)]
= (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. ∆ = $\left|\begin{array}{lll} x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y \end{array}\right|$.
Operating R₁ → R₁ – R₂, R₂ – R₃, we get:

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)

Expanding with the help of elements of R₁, we get
∆ = (5ac + 4)(x – 4)²[(1 – 0)]
= (5x + 4)(x – 4)2 = R.H.S.

(ii)

Question 11.

Solution:

Question 12.
$\left|\begin{array}{lll} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|$ = (1 – x³)²
Solution:
L.H.S = ∆ = $\left|\begin{array}{lll} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|$
Operating C₁ → C₁ + C₂ + C₃, we get:

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 + b)³.
Solution:

Expanding with the help of elements of first row, we get
∆ = (1 + a² + b²)²[(1 – a² – b²) + 2a² + 2b²]
= (1 + a² + b²)² (1 + a² + b²)
= (1 + a² + b²)³ = R.H.S.

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:
L.H.S = ∆ = $\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|$
= $\left|\begin{array}{lll} a^{2}+1 & a b+0 & a c+0 \\ a b+0 & b^{2}+1 & b c+0 \\ c a+0 & c b+0 & c^{2}+1 \end{array}\right|$
This may be expressed as the sum of 8 determinants as shown below:

Question 15.
Let A be a square matrix of order 3 x 3, then | kA | is equal to
(A) k|A|
(B) k²|A|
(C) k³|A|
(D) 3k|A|
Solution:

Taking out k common from each row, we get
|kA| = k³$\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|$ = k³|A|.
⇒ Part (C) is the correct answer.

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 them.
Solution:
Determinant is a number associated to a square matrix.
⇒ Part (C) is the correct.