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NCERT Solutions For Class 12 Maths Chapter 4 Determinants Ex 4.5

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TextbookNCERT
BoardCBSE
CategoryNCERT Solutions
ClassClass 12
SubjectMaths
ChapterChapter 4
ExerciseClass 12 Maths Chapter 4 Determinants Exercise 4.5
Number of Questions Solved18
NCERT Solutions For Class 12 Maths Chapter 4 Determinants Ex 4.5

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5

NCERT TEXTBOOK EXERCISES

Ex 4.5 Class 12 Maths Question 1.
$\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}=A(say)$
Solution:

Let Cij be cofactor of aij 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 = $\begin{bmatrix} 4 & -3 \ -2 & 1 \end{bmatrix}$
= $\begin{bmatrix} 4 & -3 \ -2 & 1 \end{bmatrix}$

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 cij be cofactor of aij 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 cij be cofactor of aij 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 cij be cofactor of aij 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:

$A=\begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix}$
$\begin{bmatrix} 8 & 5 \ -5 & 3 \end{bmatrix}$

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.

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.5 PDF

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