These NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.1
Question 1.
Evaluate the following determinant:
\(\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}\)
Solution:
\(\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}\)
= 2 x (- 1) – (- 5) x (4)
= – 2 + 20
= 18
Question 2.
(i) \(\begin{vmatrix} cos\theta & \quad -sin\theta \\ sin\theta & \quad cos\theta \end{vmatrix}\)
(ii) \(\begin{vmatrix} { x }^{ 2 }-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix}\)
Solution:
(i) \(\begin{vmatrix} cos\theta & \quad -sin\theta \\ sin\theta & \quad cos\theta \end{vmatrix}\)
= cosθ cosθ – (sinθ)(-sinθ)
= cos²θ + sin²θ
= 1
(ii) \(\begin{vmatrix} { x }^{ 2 }-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix}\)
= (x² – x + 1) (x + 1) – (x + 1) (x – 1)
= x³ – x² + x + x² – x + 1 – x² + 1
= x³ – x² + 2
Question 3.
If \(A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\) then show that |2A|=|4A|
Solution:
\(A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\)
⇒ \(2A=\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}\)
L.H.S = |2A|
= \(2A=\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}\)
= – 24
4|A| = 4|\(\left|\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right|\)| = 4(2 – 8) = 4 x – 6 = – 24
∴ |2A| = 4|A|
Question 4.
\(A=\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right] \) , then show that |3A| = 27|A|
Solution:
Question 5.
Evaluate the following determinant:
(i) \(\left| \begin{matrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{matrix} \right| \)
(ii) \(\left| \begin{matrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{matrix} \right| \)
(iii) \(\left| \begin{matrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{matrix} \right| \)
(iv) \(\left| \begin{matrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{matrix} \right| \)
Solution:
(i) \(\left| \begin{matrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{matrix} \right| \)
Question 6.
If \(\left[ \begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{matrix} \right] \), find |A|
Solution:
|A| = \(\left[ \begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{matrix} \right] \)
= 1(-9+12)-1(-18+15)-2(8-5)
= 0
Question 7.
Find the values of x, if
(i) \(\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
(ii)\(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}\)
Solution:
(i) \(\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
⇒ 2 – 20 = 2x² – 24
⇒ x² = 3
⇒ x = ±\(\sqrt{3}\)
(ii) \(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}\)
or
2 × 5 – 4 × 3 = 5 × x – 2x × 3
⇒ x = 2
Question 8.
If \(\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}\), then x is equal to
(a) 6
(b) +6
(c) -6
(d) 0
Solution:
(b) \(\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}\)
⇒ x² – 36 = 36 – 36
⇒ x² = 36
⇒ x = ± 6