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

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

NCERT TEXTBOOK EXERCISES

Ex 4.1 Class 12 Maths Question 1.
Evaluate the following determinant:
$\begin{vmatrix} 2 & 4 \ -5 & -1 \end{vmatrix}$

Solution:

$\begin{vmatrix} 2 & 4 \ -5 & -1 \end{vmatrix}$
= 2x(-1)-(-5)x(4)
=-2+20
=18

Ex 4.1 Class 12 Maths 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

Ex 4.1 Class 12 Maths 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

Ex 4.1 Class 12 Maths 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:

3A = $3\left[ \begin{matrix} 1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4 \end{matrix} \right]$
= $3\left[ \begin{matrix} 3 & 0 & 3 \ 0 & 3 & 6 \ 0 & 0 & 12 \end{matrix} \right]$

Ex 4.1 Class 12 Maths 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) $\left| \begin{matrix} 3 & -1 & -2 \ 0 & 0 & -1 \ 3 & -5 & 0 \end{matrix} \right|$

Ex 4.1 Class 12 Maths 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

Ex 4.1 Class 12 Maths 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 = ±√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

Ex 4.1 Class 12 Maths 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

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

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