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.