NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5

Get Free NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Determinants Exercise 4.5 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 4 Class 12 Determinants Ex 4.5 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Determinants chapter are the following:

Section Name	Topic Name
4	Determinants
4.1	Introduction
4.2	Determinant
4.3	Properties of Determinants
4.4	Area of a Triangle
4.5	Adjoint and Inverse of a Matrix
4.6	Applications of Determinants and Matrices
4.7	Summary

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5

Find the adjoint of each of the matrices in Questions 1 and 2.

Ex 4.5 Class 12 Maths Question 1.
$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}=A(say)$
Solution:
Let C_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by
C₁₁ = (-1)¹⁺¹ (4) = 4; C₁₂ = (-1)¹⁺²(3) = -3
C₂₁ = (-1)²⁺¹ (2)= – 2; C₂₂ = (-1)²⁺² (1) = -1
Adj A = $\begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}$
= $\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$

Ex 4.5 Class 12 Maths Question 2.
$\left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{matrix} \right] =A(say)$
Solution:
${ A }_{ 11 }={ (-1) }^{ 1+1 }M_{ 11 }=\begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix}=3$
Similarly,
NCERT Solutions for Class 12 Maths Chapter 4 Determinants 2

Verify A (adjA) = (adjA) •A = |A| I in Qs. 3 and 4.
Ex 4.5 Class 12 Maths Question 3.
$\begin{bmatrix} 2 & 3 \\ -4 & 6 \end{bmatrix}=A(say)$
Solution:
|A| = 24
NCERT Solutions for Class 12 Maths Chapter 4 Determinants 3

Ex 4.5 Class 12 Maths Question 4.
$\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right] =A(say)$
Solution:
A₁₁ = 0, A₁₂= – 11, A₁₃= 0,
A₂₁= – 3, A₂₂= 1, A₂₃= 1, A₃₁ = – 2
A₃₂= 8, A₃₃= – 3
NCERT Solutions for Class 12 Maths Chapter 4 Determinants 4

Find the inverse of each of the matrices (if it exists) given in Questions 5 to 11:

Ex 4.5 Class 12 Maths Question 5.
$\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}=A(say)$
Solution:
$\left| A \right| =\begin{vmatrix} 2 & -2 \\ 4 & 3 \end{vmatrix}=6+8=14\neq 0$
So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by
vedantu class 12 maths Chapter 4 Determinants 5

Ex 4.5 Class 12 Maths Question 6.
$\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}=A(say)$
Solution:
$\left| A \right| =\begin{vmatrix} -1 & 5 \\ -3 & 2 \end{vmatrix}=-2+15=13\neq 0$
So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by
NCERT Solutions for Class 12 Maths Chapter 4 Determinants 6

Ex 4.5 Class 12 Maths Question 7.
$\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{matrix} \right] =A$
Solution:
|A| = 10
$\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{matrix} \right] =A$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants 7

Ex 4.5 Class 12 Maths Question 8.
$\left[ \begin{matrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{matrix} \right] =A$
Solution:
$\left| A \right| =\left| \begin{matrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{matrix} \right| =1\left| \begin{matrix} 3 & 0 \\ 2 & -1 \end{matrix} \right| =-3\neq 0$
So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ijin A. Then, the cofactors of elements of A are given by
NCERT Solutions for Class 12 Maths Chapter 4 Determinants 8

Ex 4.5 Class 12 Maths Question 9.
$\left[ \begin{matrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{matrix} \right] =A$
Solution:
|A| = $\left[ \begin{matrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{matrix} \right] =A$
= 2(-1-0)-1(4-0)+3(8-3)
So, A is non-singular matrix and therefore, it is invertible.
vedantu class 12 maths Chapter 4 Determinants 9

Ex 4.5 Class 12 Maths Question 10.
$\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] =A$
Solution:
|A| = $\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] =A$
= 1(8-6)+1(0+9)+2(0-6)
= 2+9-12
= -1≠0
∴A is invertible and
${ A }^{ -1 }=\frac { Adj\quad A }{ |A| }$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants 10

Ex 4.5 Class 12 Maths Question 11.
$\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & cos\alpha & sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{matrix} \right]$
Solution:
A = $\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & cos\alpha & sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{matrix} \right]$
adj A = $\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -cos\alpha & -sin\alpha \\ 0 & -sin\alpha & cos\alpha \end{matrix} \right]$
First find |A| = -cos²α-sin²α
=-1≠0
vedantu class 12 maths Chapter 4 Determinants 11

Ex 4.5 Class 12 Maths Question 12.
Let $A=\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix},B=\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$ , verify that (AB)^-1 = B^-1A^-1
Solution:
Here |A| = $A=\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$
= 15-14
= 1≠0
$Adj A=\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants 12

Ex 4.5 Class 12 Maths Question 13.
If $A=\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ show that A² – 5A + 7I = 0,hence find A^-1
Solution:
A = $\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$
A² = $\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants 13

Ex 4.5 Class 12 Maths Question 14.
For the matrix A = $\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$ find file numbers a and b such that A²+aA+bI²=0. Hence, find A^-1.
Solution:
A = $\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$
A²+aA+bI²=0
NCERT Solutions for Class 12 Maths Chapter 4 Determinants 14

Ex 4.5 Class 12 Maths Question 15.
For the matrix $A=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{matrix} \right]$ Show that A³-6A²+5A+11I₃=0.Hence find A^-1
Solution:
A² = $\left[ \begin{matrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants 15
vedantu class 12 maths Chapter 4 Determinants 15.1

Ex 4.5 Class 12 Maths Question 16.
If $A=\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right]$ show that A³-6A²+9A-4I=0 and hence, find A^-1
Solution:
We have
$A=\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants 16

Ex 4.5 Class 12 Maths Question 17.
Let A be a non-singular square matrix of order 3×3. Then | Adj A | is equal to:
(a) | A |
(b) | A |²
(c) | A |³
(d) 3 | A |
Solution:
Let A = $\left[ \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants 17

Dividing by | A |, |Adj. A| = | A |²
Hence, Part (b) is the correct answer.

Ex 4.5 Class 12 Maths Question 18.
If A is an invertible matrix of order 2, then det. (A^-1) is equal to:
(a) det. (A)
(b) $\\ \frac { 1 }{ det.(A) }$
(c) 1
(d) 0
Solution:
|A|≠0
=> A^-1 exists => AA^-1 = I
|AA^-1| = |I| = I
=> |A||A^-1| = I
$|{ A }^{ -1 }|=\frac { 1 }{ |A| }$
Hence option (b) is correct.