NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths 4 Determinants Ex 4.5.

• Determinants Class 12 Ex 4.1
• Determinants Class 12 Ex 4.2
• Determinants Class 12 Ex 4.3
• Determinants Class 12 Ex 4.4
• Determinants Class 12 Ex 4.6

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 4
Chapter Name	Determinants
Exercise	Ex 4.5
Number of Questions Solved	18
Category	NCERT Solutions

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,

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

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

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

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

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
\(\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{matrix} \right] =A\)

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_ij in A. Then, the cofactors of elements of A are given by

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.

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
\({ A }^{ -1 }=\frac { Adj\quad A }{ |A| } \)

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

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
\(Adj A=\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} \)

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 = \(\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \)

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

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 = \( \left[ \begin{matrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{matrix} \right] \)

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
\(A=\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right] \)

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 = \(\left[ \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right] \)

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.

 

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