Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 4 Determinants Ex 4.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 4 Determinants Ex 4.1

Question 1.
(left|begin{array}{cc} 2 & 4 \ -5 & -1 end{array}right|)
Solution:
(left|begin{array}{cc} 2 & 4 \ -5 & -1 end{array}right|) = 2 x (-1) – 4 x (-5) = – 2 + 20 = 18

Question 2.
(i) (left|begin{array}{cc} cos theta & -sin theta \ sin theta & cos theta end{array}right|)
(ii) (left|begin{array}{cc} x^{2}-x+1 & x-1 \ x+1 & x+1 end{array}right|)
Solution:
(i) (left|begin{array}{cc} cos theta & -sin theta \ sin theta & cos theta end{array}right|)
= cos θ x cos θ – sin θ x (- sin θ)
= cos²θ + sin²θ = 1.

(ii) (left|begin{array}{cc} x^{2}-x+1 & x-1 \ x+1 & x+1 end{array}right|)
= (x² – x + 1)(x + 1) – (x – 1)(x + 1)
= (x³ + 1) – (x² – 1)
= x³ + 1 – x² + 1 = x³ – x² + 2.

Question 3.
Solution:
If A = (left[begin{array}{ll} 1 & 2 \ 4 & 2 end{array}right]), then show that | 2A | = 4| A |
Solution:
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 | = (left|begin{array}{cc} 2 & 4 \ 8 & 4 end{array}right|) = 2 x 2(left|begin{array}{cc} 1 & 2 \ 4 & 2 end{array}right|) = 4| A |.
[ Taking out 2 from C₁ and C₂ each.]

Question 4.
If A = (left[begin{array}{lll} 1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4 end{array}right]), then show that | 3A | = 27 | A |.
Solution:

Note: When a matrix is multipled by k, each element of matrix is multipled by k, whereas in case of determinant, only elements of a row or a column are multiplied by k.

Question 5.
Evaluate the determinants:
(i) (left|begin{array}{ccc} 3 & -1 & -2 \ 0 & 0 & -1 \ 3 & -5 & 0 end{array}right|)
(ii) (left|begin{array}{ccc} 3 & -4 & 5 \ 1 & 1 & -2 \ 2 & 3 & 1 end{array}right|)
(iii) (left|begin{array}{ccc} 0 & 1 & 2 \ -1 & 0 & -3 \ -2 & 3 & 0 end{array}right|)
(iv) (left|begin{array}{ccc} 2 & -1 & -2 \ 0 & 2 & -1 \ 3 & -5 & 0 end{array}right|)
Solution:
(i)

(ii) (left|begin{array}{ccc} 3 & -4 & 5 \ 1 & 1 & -2 \ 2 & 3 & 1 end{array}right|)
= 3 x [1 x 1 – (- 2) x 3] + 4[1 x 1 – (- 2) x 2] + 5[1 x 3 – 2 x 1]
= 3(1 + 6) + 4(1 + 4) + 5(3 – 2)
= 3 x 7 + 4 x 5 + 5 x 1
= 21 + 20 + 5 = 46.

(iii) (left|begin{array}{ccc} 0 & 1 & 2 \ -1 & 0 & -3 \ -2 & 3 & 0 end{array}right|)
= (0 + 9) – 1.[- 1 x 0 – (- 2)(- 3)] + 2 [(- 1) x 3 – (- 2) x 0]
= 0 – (- 6) + 2(- 3) = 6 – 6 = 0.

(iv) Expanding with the help of elements of I column,
(left|begin{array}{ccc} 2 & -1 & -2 \ 0 & 2 & -1 \ 3 & -5 & 0 end{array}right|) = 2(left|begin{array}{cc} 2 & -1 \ 5 & 0 end{array}right|) – 0 x (left|begin{array}{cc} -1 & -2 \ -5 & 0 end{array}right|) + 3(left|begin{array}{cc} -1 & -2 \ 2 & -1 end{array}right|)
= 2(0 – 5) – 0 + 3(1 + 4) = – 10 + 15 = 5.

Question 6.
If A = (left|begin{array}{ccc} 1 & 1 & -2 \ 2 & 1 & -3 \ 5 & 4 & -9 end{array}right|), find | A |.
Solution:
| A | = (left|begin{array}{ccc} 1 & 1 & -2 \ 2 & 1 & -3 \ 5 & 4 & -9 end{array}right|)
= 1.(left|begin{array}{cc} 1 & -3 \ 4 & -9 end{array}right|) – 1.(left|begin{array}{cc} 2 & -3 \ 5 & -9 end{array}right|) – 2.(left|begin{array}{cc} 2 & 1 \ 5 & 4 end{array}right|)
= (- 9 + 12) – ( – 18 + 15) – 2(8 – 5)
= 3 + 3 – 6 = 0.

Question 7.
Find the value of x, if
(i) (left|begin{array}{cc} 2 & 4 \ 5 & 1 end{array}right|) = (left|begin{array}{cc} 2x & 4 \ 6 & x end{array}right|)
(ii) (left|begin{array}{cc} 2 & 3 \ 4 & 5 end{array}right|) = (left|begin{array}{cc} x & 3 \ 2x & 5 end{array}right|)
Solution:
(i) (left|begin{array}{cc} 2 & 4 \ 5 & 1 end{array}right|) = (left|begin{array}{cc} 2x & 4 \ 6 & x end{array}right|)
or 2 – 20 = (2x² – 24)
or – 18 = 2x² – 24
or 2x² = 6
x = ±(sqrt{3})

(ii) (left|begin{array}{cc} 2 & 3 \ 4 & 5 end{array}right|) = (left|begin{array}{cc} x & 3 \ 2x & 5 end{array}right|)
or 2 x 5 – 4 x 3 = 5 × x – 2x × 3
or 10 – 12 = 5x – 6x
or – 2 = – x
∴ x = 2.

Question 8.
If (left|begin{array}{cc} x & 2 \ 18 & x end{array}right|) = (left|begin{array}{cc} 6 & 2 \ 18 & 6 end{array}right|), then x is equal to
(A) 6
(B) ± 6
(C) – 6
(D) 6, 6
Solution:
(left|begin{array}{cc} x & 2 \ 18 & x end{array}right|) = (left|begin{array}{cc} 6 & 2 \ 18 & 6 end{array}right|)
or x² – 18 x 2 = 36 – 36
or x² – 36 = 0
x² = 36
x² = ± 6.
∴ Part (B) is the correct answer.