# NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1

These NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-4-ex-4-1/

## NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.1

Class 12 Maths Chapter 4 Exercise 4.1 Solutions 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

12th Maths Chapter 4 Exercise 4.1 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

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