# 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 â†’ 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 â†’ 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 â†’ 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 â†’ 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.
$$\left|\begin{array}{ccc} -a^{2} & a b & a c \\ b a & -b^{2} & b c \\ a c & c b & -c^{2} \end{array}\right|$$ = 4aÂ²bÂ²cÂ².
Solution:
Taking a, b, c common from I, II and III rows respectively,
we get âˆ† = $$\left|\begin{array}{ccc} -a & b & c \\ a & -b & c \\ a & b & -c \end{array}\right|$$
Again a, b and c are taking out common from I, II and III columns respectively. we get
âˆ´ âˆ† = aÂ²bÂ²cÂ²$$\left|\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right|$$
Now operating R1 â†’ R1 + R2, we get:
âˆ† = aÂ²bÂ²cÂ²$$\left|\begin{array}{ccc} 0 & 0 & 2 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right|$$ = aÂ²bÂ²cÂ².2(1+1) = 4aÂ²bÂ²cÂ².

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,
âˆ† = (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
âˆ† = (a-b)(b-c)$$\left|\begin{array}{ll} 1 & a+b \\ 1 & b+c \end{array}\right|$$
â‡’ âˆ† = (a – b)(b – c)[(b + c) – (a + b)]
â‡’ âˆ† = (a – b)(b – c)(c – a)

(ii)

Taking out a – b and b – c common from C1 and C2 respectively,
âˆ† = (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 + cÂ²) – (aÂ² + ab + bÂ²)]
= (a – b)(b – c)[(bc – ab) + (cÂ² – aÂ²)]
= (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. âˆ† = $$\left|\begin{array}{lll} x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y \end{array}\right|$$.
Operating R1 â†’ 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)Â²
(ii) $$\left|\begin{array}{ccc} y+k & y & y \\ y & y+k & y \\ y & y & y+k \end{array}\right|$$ = kÂ²(3y + k)
Solution:
(i)

Expanding with the help of elements of R1, we get
âˆ† = (5ac + 4)(x – 4)Â²[(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 – xÂ³)Â²
Solution:
L.H.S = âˆ† = $$\left|\begin{array}{lll} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|$$
Operating C1 â†’ 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 + aÂ² + b)Â³.
Solution:

Expanding with the help of elements of first row, we get
âˆ† = (1 + aÂ² + bÂ²)Â²[(1 – aÂ² – bÂ²) + 2aÂ² + 2bÂ²]
= (1 + aÂ² + bÂ²)Â² (1 + aÂ² + bÂ²)
= (1 + aÂ² + bÂ²)Â³ = 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 + aÂ² + bÂ² + cÂ².
Solution:
L.H.S = âˆ† = $$\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|$$
= $$\left|\begin{array}{lll} a^{2}+1 & a b+0 & a c+0 \\ a b+0 & b^{2}+1 & b c+0 \\ c a+0 & c b+0 & c^{2}+1 \end{array}\right|$$
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) kÂ²|A|
(C) kÂ³|A|
(D) 3k|A|
Solution:

Taking out k common from each row, we get
|kA| = kÂ³$$\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|$$ = kÂ³|A|.
â‡’ 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.
â‡’ Part (C) is the correct.