Can I prepare only using this material?

Yes, RD Sharma Class 12th Exercise 6.2 is designed to cater to students’ needs and is the best material to prepare for exams.

Do they have solutions for other books?

Yes, Career360 has many solutions for different classes and books on their website.

What is a Determinant?

Determinant of a matrix is a singular value obtained using the values of the square matrix. For more details, check out Class 12 RD Sharma Chapter 6 Exercise 6.2 Solution.

Can I use this material for entrance exams?

Yes, RD Sharma Class 12 Solutions Adjoint and Inverse of Matrix Ex 6.2 is best suited for exam preparation.

Are there any hidden charges?

RD Sharma Class 12 Solutions Chapter 6 Ex 6.2 is absolutely free to download with no hidden costs.

RD Sharma Class 12 Exercise 6.2 Adjoint and Inverse of Matrix Solutions Maths - Download PDF Free Online

Schools
RD Sharma Solutions
RD Sharma Class 12 Exercise 6.2 Adjoint and Inverse of Matrix Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 6.2 Adjoint and Inverse of Matrix Solutions Maths - Download PDF Free Online

Updated on 20 Jan 2022, 03:00 PM IST

RD Sharma books are considered as one of the best materials for maths. They are widely used in the country by many CBSE schools and are the number one choice of teachers. RD Sharma provides the most explicit material, so it is used as the default maths textbook in many schools.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter6 Adjoint & Inverse of Matrix - Other Exercise

Adjoint and Inverse of Matrix excercise 6.2 question 1

Answer:
$\frac{1}{25}\begin{bmatrix} 3 &1 \\ 4&-7 \end{bmatrix}$
Hint: Here we use the concept of elementary row operation
Given: $\begin{bmatrix} 7 &1 \\ 4&-3 \end{bmatrix}$
Solution: $A = IA$
$A = \begin{bmatrix} 1& 0\\ 0& 1 \end{bmatrix}A$
$\Rightarrow \begin{bmatrix} 7& 1\\ 4& -3 \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}A$
Applying $R_1 \rightarrow \frac{1}{7}R_1$
$\Rightarrow \begin{bmatrix} 1 &\frac{1}{7} \\ 4&-3 \end{bmatrix} =\begin{bmatrix} \frac{1}{7} &0 \\ 0&1 \end{bmatrix}A$

Applying $R_2\rightarrow R_2-4R_1$
$\Rightarrow \begin{bmatrix} 1 &\frac{1}{7} \\ 0& \frac{-25}{7} \end{bmatrix}$$=\begin{bmatrix} \frac{1}{7} &0 \\ \frac{-4}{7}&1 \end{bmatrix}A$
Applying $R_2\rightarrow \frac{-7}{25}R_2$
$\Rightarrow \begin{bmatrix} 1 & \frac{1}{7}\\ 0&1 \end{bmatrix} =\begin{bmatrix} \frac{1}{7}&0 \\ \frac{4}{25}&\frac{-7}{25} \end{bmatrix}$
Applying $R_1\rightarrow R_1-\frac{1}{7}R_1$
$\Rightarrow \begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix} =\begin{bmatrix} \frac{21}{75}&\frac{1}{25} \\ \frac{4}{25}&\frac{-7}{25} \end{bmatrix}A$
$\Rightarrow \begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix} =\begin{bmatrix} \frac{3}{25}&\frac{1}{25} \\ \frac{4}{25}&\frac{-7}{25} \end{bmatrix}A$
$\Rightarrow A^{-1} = \frac{1}{25}\begin{bmatrix} 3 & 1\\ 4& -7 \end{bmatrix}$

Adjoint and Inverse of Matrix excercise 6.2 question 2

Answer: $A^{-1}=\begin{bmatrix} 1 &-2 \\ -2& 5 \end{bmatrix}$
Hint: Here, we use the basic of matrix transpose
Given: $\begin{bmatrix} 5 &2 \\ 2& 1 \end{bmatrix}$
Solution: Let $A = IA$
$A = \begin{bmatrix} 5 &2 \\ 2& 1 \end{bmatrix}, I = \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} 5 &2 \\ 2& 1 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0& 1 \end{bmatrix}A$

Applying $R_1 = \frac{1}{5}R_1$
$\Rightarrow \begin{bmatrix} 1 &\frac{2}{5} \\ 0&\frac{1}{5} \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & 0\\ 0& 1 \end{bmatrix}A$

Applying $R_2 = R_2 - 2R_1$

$\Rightarrow \begin{bmatrix} 1 &\frac{2}{5} \\ 0&\frac{1}{5} \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & 0\\ \frac{-2}{5}& 1 \end{bmatrix}A$
$\Rightarrow \begin{bmatrix} 1 & \frac{2}{5}\\ 0&1 \end{bmatrix}=\begin{bmatrix} \frac{1}{5}&0 \\ -2& 5 \end{bmatrix}A$
Applying $R_2\rightarrow 5R_2$

$\Rightarrow \begin{bmatrix} 1 &\frac{2}{5} \\ 0 &1 \end{bmatrix}=\begin{bmatrix} \frac{1}{5} &0 \\ -2& 5 \end{bmatrix}A$
Applying $R_1 \rightarrow R_1-\frac{2}{5}R_2$

$\Rightarrow \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}=\begin{bmatrix} 1 &-2 \\ -2& 5 \end{bmatrix}A$

So, Here $A^{-1}=\begin{bmatrix} 1 &-2 \\ -2& 5 \end{bmatrix}$

Adjoint and Inverse of Matrix excercise 6.2 question 4

Answer:$A^{-1}=\left[\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right]$
Hint: Here, we use the concept of matrix inverse using elementary row operation
Given: $\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]$
Solution: Let $A = \left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]$
$A = IA$
$A=\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right], I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
$\Rightarrow\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A$
Applying $R_{1} \rightarrow R_{1}-R_{2}$
$\Rightarrow\left[\begin{array}{ll} 1 & 2 \\ 1 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right] A$
Applying $R_{2} \rightarrow R_{2}-R_{1}$
$\Rightarrow\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ -1 & 2 \end{array}\right] A$
Applying $R_{1} \rightarrow R_{1}-2R_{2}$
$\Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right] A$

Hence, $A^{-1}=\left[\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right]$

Adjoint and Inverse of Matrix excercise 6 point 2 question 5

Answer: $\left[\begin{array}{cc} 7 & -10 \\ -2 & 3 \end{array}\right]$
Hint: Here, we use the concept of matrix inverse using elementary row operation
Given: $A=\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right]$
Solution: Let $A=\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right]$
$A = IA$
$A=\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right], I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
$\Rightarrow\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A$
Applying $R_{1} \rightarrow \frac{1}{3} R_{1}$
$\Rightarrow\left[\begin{array}{ll} 1 & \frac{10}{3} \\ 2 & 7 \end{array}\right]=\left[\begin{array}{ll} \frac{1}{3} & 0 \\ 0 & 1 \end{array}\right] A$
Applying $R_{2} \rightarrow R_{2}-2 R_{1}$
$\Rightarrow\left[\begin{array}{ll} 1 & \frac{10}{3} \\ 0 & \frac{1}{3} \end{array}\right]=\left[\begin{array}{cc} \frac{1}{3} & 0 \\ \frac{-2}{3} & 1 \end{array}\right] A$

Applying $R_2\rightarrow 3R_2$
$\Rightarrow\left[\begin{array}{ll} 1 & \frac{10}{3} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} \frac{1}{3} & 0 \\ -2 & 3 \end{array}\right] A$
Applying $R_{1} \rightarrow R_{1}-\frac{10}{3} R_{2}$
$\Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 7 & -10 \\ -2 & 3 \end{array}\right]$

Hence, $\left[\begin{array}{cc} 7 & -10 \\ -2 & 3 \end{array}\right]$

Adjoint and Inverse of Matrix excercise 6.2 question 6

Answer: $\left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{array}\right]$
Hint: Here, we use the concept of matrix inverse using elementary row operation
Given: $\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]$
Solution: Let $A=\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]$
$A = IA$
$A=\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right], I=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
$\Rightarrow\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying $R_{1} \leftrightarrow R_{2}$
$\Rightarrow\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1 \end{array}\right]=\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying $R_{3} \rightarrow R_{3}-3 R_{1}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{array}\right] A$
Applying $R_{1} \rightarrow R_{1}-2 R_{2}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{array}\right]=\left[\begin{array}{ccc} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{array}\right] A$
Applying $R_{3} \rightarrow R_{3}+5 R_{2}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{array}\right]=\left[\begin{array}{ccc} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1 \end{array}\right] A$
Applying $R_{3} \rightarrow \frac{1}{2} R_{3}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -2 & 1 & 0 \\ 1 & 0 & 0 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{array}\right] A$
Applying $R_{1} \rightarrow R_{1}+R_{3}$
$\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ 1 & 0 & 0 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{array}\right] A$
Applying $R_{2} \rightarrow R_{2}-2 R_{3}$
$\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{array}\right] A$
So,$\left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{array}\right]$

Adjoint and Inverse of Matrix excercise 6.2 question 7

Answer: $\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right]$

Hint: Here, we use the concept of matrix inverse using elementary row operation

Given: $\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]$

Solution: Let $A=\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]$

$A = IA$
$\begin{aligned} &A=\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right], I=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{lll} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A \end{aligned}$

Applying $R_{1} \rightarrow \frac{1}{2} R_{1}$

$\Rightarrow\left[\begin{array}{lll} 1 & 0 & -\frac{1}{2} \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{lll} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$

Applying $R_{2} \rightarrow R_{2}-5 R_{1}$

$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{lll} \frac{1}{2} & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$

Applying $R_{3} \rightarrow R_{3}-R_{2}$

$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ \frac{5}{2} & -1 & 1 \end{array}\right] A$

Applying $R_{3} \rightarrow 2 R_{3}$

$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ 5 & -2 & 2 \end{array}\right] A$

Applying

$\begin{aligned} &R_{1} \rightarrow R_{1}+\frac{1}{2} R_{3} \ \\ &R_{2} \rightarrow R_{2}-\frac{5}{2} R_{3} \\ &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] A \end{aligned}$

So,$A^{-1}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right]$

Adjoint and inverse of matrix exercise 6 point 2 question 8

Answer: $\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & 1 & 0 \\ 2 & -5 & 2 \end{array}\right]$

Hint: Here, we use the concept of matrix inverse using elementary row operation

Given: $\left[\begin{array}{lll} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{array}\right]$

Solution: Let $A=\left[\begin{array}{lll} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{array}\right]$

$A = IA$

$\begin{aligned} &A=\left[\begin{array}{lll} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{array}\right], I=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{lll} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A \end{aligned}$

Applying $R_{1} \rightarrow \frac{1}{2} R_{1}$

$\Rightarrow\left[\begin{array}{lll} 1 & \frac{3}{2} & \frac{1}{2} \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{array}\right]=\left[\begin{array}{lll} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$

Applying

$\begin{aligned} & R_{2} \rightarrow R_{2}-2 R_{1} \\ R_{3} & \rightarrow R_{3}-3 R_{1} \\ \Rightarrow\left[\begin{array}{lll} 1 & \frac{3}{2} & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & \frac{5}{2} & \frac{1}{2} \end{array}\right] &=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -1 & 1 & 0 \\ \frac{-3}{2} & 0 & 1 \end{array}\right] A \end{aligned}$

Applying

$\begin{aligned} &R_{1} \rightarrow R_{1}-\frac{3}{2} R_{2} \\ &R_{3} \rightarrow R_{3}-\frac{5}{2} R_{2} \\ &\Rightarrow\left[\begin{array}{lll} 1 & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} 2 & \frac{-3}{2} & 0 \\ -1 & 1 & 0 \\ 1 & \frac{-5}{2} & 1 \end{array}\right] A \end{aligned}$

Applying $R_{3} \rightarrow 2 R_{3}$

$\Rightarrow\left[\begin{array}{lll} 1 & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 2 & \frac{-3}{2} & 0 \\ -1 & 1 & 0 \\ 2 & -5 & 2 \end{array}\right] A$

Applying $R_{1} \rightarrow R_{1}-\frac{1}{2} R_{3}$

$\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & 1 & 0 \\ 2 & -5 & 2 \end{array}\right] A$

So,$A^{-1}=$ $\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & 1 & 0 \\ 2 & -5 & 2 \end{array}\right]$

Adjoint and Inverse of Matrix excercise 6.2 question 9

Answer: $\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right]$
Hint: Here, we use the concept of matrix inverse using elementary row operation
Given: $\left[\begin{array}{rrr} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]$
Solution: Let $A=\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]$
$\begin{aligned} &A=I A \\ &A=\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right], I=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A \end{aligned}$

Applying $R_{1} \rightarrow \frac{1}{3} R_{1}$
$\Rightarrow\left[\begin{array}{lll} 1 & -1 & \frac{4}{3} \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]=\left[\begin{array}{lll} \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying $R_{2} \rightarrow R_{2}-2 R_{1}$
$\Rightarrow\left[\begin{array}{ccc} 1 & -1 & \frac{4}{3} \\ 0 & -1 & \frac{4}{3} \\ 0 & -1 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{-2}{3} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying $R_{2} \rightarrow-R_{2}$
$\Rightarrow\left[\begin{array}{ccc} 1 & -1 & \frac{4}{3} \\ 0 & 1 & \frac{-4}{3} \\ 0 & -1 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{2}{3} & -1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying
$\begin{aligned} &R_{1} \rightarrow R_{1}+R_{2} \\ &R_{3} \rightarrow R_{3}+R_{2} \\ &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & \frac{-4}{3} \\ 0 & 0 & \frac{-1}{3} \end{array}\right]=\left[\begin{array}{ccc} 1 & -1 & 0 \\ \frac{2}{3} & -1 & 0 \\ \frac{2}{3} & -1 & 1 \end{array}\right] A \end{aligned}$

Applying $R_{3} \rightarrow-3 R_{3}$
$\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right] A$
Applying $R_{2} \rightarrow R_{2}+\frac{4}{3} R_{3}$
$\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right] A$

So, $A^{-1}=\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right]$

Adjoint and Inverse of Matrix excercise 6 point 2 question 2

Answer: $\left[\begin{array}{ccc} \frac{-4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & \frac{-1}{2} & \frac{-1}{6} \\ \frac{5}{6} & \frac{-1}{2} & \frac{1}{6} \end{array}\right]$
Hint: Here, we use the concept of matrix inverse using elementary row operation
Given: $\left[\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 3 & -1 \\ 1 & -1 & 3 \end{array}\right]$
Solution: Let $A=\left[\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 3 & -1 \\ 1 & -1 & 3 \end{array}\right]$
$\begin{aligned} &A=I A \\ &A=\left[\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 3 & -1 \\ 1 & -1 & 3 \end{array}\right], I=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 3 & -1 \\ 1 & -1 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A \end{aligned}$
Applying
$\begin{gathered} R_{2} \rightarrow R_{2}-2 R_{1} \\ R_{3} \rightarrow R_{3}-R_{1} \\ \Rightarrow\left[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & -1 & -1 \\ 0 & -3 & 3 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -2 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right] A \end{gathered}$
Applying $R_{2} \rightarrow-R_{2}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & -3 & 3 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 2 & -1 & 0 \\ -1 & 0 & 1 \end{array}\right] A$
Applying
$\begin{aligned} &R_{1} \rightarrow R_{1}-2 R_{2} \\ &R_{3} \rightarrow R_{3}+3 R_{2} \\ &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 6 \end{array}\right]=\left[\begin{array}{ccc} -3 & 2 & 0 \\ 2 & -1 & 0 \\ 5 & -3 & 1 \end{array}\right] A \end{aligned}$
Applying $R_{3} \rightarrow \frac{1}{6} R_{3}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -3 & 2 & 0 \\ 2 & -1 & 0 \\ \frac{5}{6} & -\frac{1}{2} & \frac{1}{6} \end{array}\right] A$
Applying
$\begin{aligned} &R_{1} \rightarrow R_{1}+2 R_{3} \\ &R_{2} \rightarrow R_{2}-R_{3} \\ &\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -\frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & \frac{-1}{2} & \frac{-1}{6} \\ \frac{5}{6} & -\frac{1}{2} & \frac{1}{6} \end{array}\right] A \end{aligned}$
So,$A^{-1}=\left[\begin{array}{ccc} -\frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & \frac{-1}{2} & \frac{-1}{6} \\ \frac{5}{6} & -\frac{1}{2} & \frac{1}{6} \end{array}\right]$

Adjoint and inverse of matrix exercise 6.2 question 11

Answer: $\left[\begin{array}{ccc} \frac{1}{15} & \frac{-2}{15} & \frac{-1}{3} \\ \frac{-11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & \frac{-1}{6} \end{array}\right]$
Hint: Here, we use the concept of matrix inverse using elementary row operation
Given: $\left[\begin{array}{ccc} 2 & -1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{array}\right]$
Solution:
Let,
$\begin{aligned} &A=\left[\begin{array}{ccc} 2 & -1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{array}\right] \\ &A=I A \\ &A=\left[\begin{array}{ccc} 2 & -1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{array}\right], I=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{lll} 2 & -1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A \end{aligned}$
Applying $R_{1} \rightarrow \frac{1}{2} R_{1}$
$\Rightarrow\left[\begin{array}{ccc} 1 & -\frac{1}{2} & \frac{3}{2} \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{array}\right]=\left[\begin{array}{lll} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying
$\begin{aligned} &R_{2} \rightarrow R_{2}-R_{1} \\ &R_{3} \rightarrow R_{3}-3 R_{1} \\ &\Rightarrow\left[\begin{array}{ccc} 1 & -\frac{1}{2} & \frac{3}{2} \\ 0 & \frac{5}{2} & \frac{5}{2} \\ 0 & \frac{5}{2} & \frac{-7}{2} \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ \frac{-1}{2} & 1 & 0 \\ \frac{-3}{2} & 0 & 1 \end{array}\right] A \end{aligned}$

Applying $R_{2} \rightarrow \frac{2}{5} R_{2}$
$\Rightarrow\left[\begin{array}{ccc} 1 & -\frac{1}{2} & \frac{3}{2} \\ 0 & 1 & 1 \\ 0 & \frac{5}{2} & \frac{-7}{2} \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ \frac{-1}{5} & \frac{2}{5} & 0 \\ \frac{-3}{2} & 0 & 1 \end{array}\right] A$
Applying
$\begin{aligned} &R_{1} \rightarrow R_{1}+\frac{1}{2} R_{2} \\ &R_{3} \rightarrow R_{3}-\frac{5}{2} R_{2} \\ &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & -6 \end{array}\right]=\left[\begin{array}{ccc} \frac{2}{5} & \frac{1}{5} & 0 \\ \frac{-1}{5} & \frac{2}{5} & 0 \\ -1 & -1 & 1 \end{array}\right] A \end{aligned}$

Applying $R_{3} \rightarrow \frac{-1}{6} R_{3}$
$\Rightarrow\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{2}{5} & \frac{1}{5} & 0 \\ \frac{-1}{5} & \frac{2}{5} & 0 \\ \frac{1}{6} & \frac{1}{6} & \frac{-1}{6} \end{array}\right] A$
Applying
$\begin{aligned} &R_{2} \rightarrow R_{2}-R_{3} \\ &R_{1} \rightarrow R_{1}-2 R_{3} \\ &\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{15} & \frac{-2}{15} & \frac{-1}{3} \\ \frac{-11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & \frac{-1}{6} \end{array}\right] A \end{aligned}$

So,$A^{-1}=\left[\begin{array}{ccc} \frac{1}{15} & \frac{-2}{15} & \frac{-1}{3} \\ \frac{-11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & \frac{-1}{6} \end{array}\right]$

Adjoint and Inverse of Matrix excercise 6.2 question 12

Answer: $\frac{1}{11}\left[\begin{array}{ccc} -2 & 5 & -1 \\ -1 & -3 & 5 \\ 7 & -1 & -2 \end{array}\right]$
Hint: Here, we use the concept of matrix inverse using elementary row operation
Given: $\left[\begin{array}{lll} 1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1 \end{array}\right]$
Solution: Let $A = IA$
$\begin{aligned} &A=\left[\begin{array}{lll} 1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1 \end{array}\right], I=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{lll} 1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A \end{aligned}$
Applying
$\begin{aligned} &R_{2} \rightarrow R_{2}-3 R_{1} \\ &R_{3} \rightarrow R_{3}-2 R_{1} \\ &\Rightarrow\left[\begin{array}{ccc} 1 & 1 & 2 \\ 0 & -2 & -5 \\ 0 & 1 & -3 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -3 & 1 & 0 \\ -2 & 0 & 1 \end{array}\right] A \end{aligned}$
Applying
$\begin{aligned} &R_{2} \rightarrow R_{2}-3 R_{1} \\ &R_{3} \rightarrow R_{3}-2 R_{1} \\ &\Rightarrow\left[\begin{array}{ccc} 1 & 1 & 2 \\ 0 & -2 & -5 \\ 0 & 1 & -3 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -3 & 1 & 0 \\ -2 & 0 & 1 \end{array}\right] A \end{aligned}$
Applying $R_{2} \rightarrow \frac{-1}{2} R_{2}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 1 & 2 \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & -3 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ \frac{3}{2} & \frac{-1}{2} & 0 \\ -2 & 0 & 1 \end{array}\right] A$
Applying
$\begin{aligned} &R_{1} \rightarrow R_{1}-R_{2} \\ &R_{3} \rightarrow R_{3}-R_{2} \\ &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & \frac{-1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & -\frac{11}{2} \end{array}\right]=\left[\begin{array}{ccc} \frac{-1}{2} & \frac{1}{2} & 0 \\ \frac{3}{2} & \frac{-1}{2} & 0 \\ -\frac{7}{2} & \frac{1}{2} & 1 \end{array}\right] A \end{aligned}$
Applying $R_{3} \rightarrow \frac{-2}{11} R_{3}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & \frac{-1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -\frac{1}{2} & \frac{1}{2} & 0 \\ \frac{3}{2} & \frac{-1}{2} & 0 \\ \frac{7}{11} & \frac{-1}{11} & \frac{-2}{11} \end{array}\right] A$
Applying , $R_{1} \rightarrow R_{1}+\frac{1}{2} R_{3}$,$R_{2} \rightarrow R_{2}-\frac{5}{2} R_{3}$
$\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -\frac{2}{11} & \frac{5}{11} & \frac{-1}{11} \\ \frac{-1}{11} & \frac{-3}{11} & \frac{5}{11} \\ \frac{7}{11} & \frac{-1}{11} & \frac{-2}{11} \end{array}\right] A$
So,$A^{-1}=\frac{1}{11}\left[\begin{array}{ccc} -2 & 5 & -1 \\ -1 & -3 & 5 \\ 7 & -1 & -2 \end{array}\right]$

Adjoint and Inverse of Matrix excercise 6.2 question 13

Answer: $\left[\begin{array}{ccc} -2 & \frac{1}{2} & 1 \\ 11 & -1 & -6 \\ 4 & \frac{-1}{2} & -2 \end{array}\right]$
Hint: Here, we use the concept of matrix inverse using elementary row operation
Given: $\left[\begin{array}{ccc} 2 & -1 & 4 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \end{array}\right]$
Solution: Let $A=\left[\begin{array}{ccc} 2 & -1 & 4 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \end{array}\right]$
$\begin{aligned} &A=I A \\ &A=\left[\begin{array}{ccc} 2 & -1 & 4 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \end{array}\right], I=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{ccc} 2 & -1 & 4 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A \end{aligned}$
Applying $R_{1} \rightarrow \frac{1}{2} R_{1}$
$\Rightarrow\left[\begin{array}{ccc} 1 & -\frac{1}{2} & 2 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying
$\begin{aligned} &R_{2} \rightarrow R_{2}-4 R_{1} \\ &R_{3} \rightarrow R_{3}-3 R_{1} \\ &\qquad \quad \Rightarrow\left[\begin{array}{ccc} 1 & -\frac{1}{2} & 2 \\ 0 & 2 & -6 \\ 0 & -\frac{1}{2} & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -2 & 1 & 0 \\ -\frac{3}{2} & 0 & 1 \end{array}\right] A \end{aligned}$

Applying $R_{2} \rightarrow \frac{1}{2} R_{2}$
$\Rightarrow\left[\begin{array}{ccc} 1 & -\frac{1}{2} & 2 \\ 0 & 1 & -3 \\ 0 & -\frac{1}{2} & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -1 & \frac{1}{2} & 0 \\ -\frac{3}{2} & 0 & 1 \end{array}\right] A$
Applying
$\begin{aligned} &R_{1} \rightarrow R_{1}+\frac{1}{2} R_{2} \\ &\begin{array}{l} R_{3} \rightarrow R_{3}+\frac{1}{2} R_{2} \\ \Rightarrow\left[\begin{array}{ccc} 1 & 0 & \frac{1}{2} \\ 0 & 1 & -3 \\ 0 & 0 & \frac{-1}{2} \end{array}\right]=\left[\begin{array}{ccc} 0 & \frac{1}{4} & 0 \\ -1 & \frac{1}{2} & 0 \\ -2 & \frac{1}{4} & 1 \end{array}\right] A \end{array} \end{aligned}$
Applying $R_{3} \rightarrow-2 R_{3}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & \frac{1}{2} \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 0 & \frac{1}{4} & 0 \\ -1 & \frac{1}{2} & 0 \\ 4 & \frac{-1}{2} & -2 \end{array}\right] A$

Applying
$\begin{aligned} &R_{1} \rightarrow R_{1}-\frac{1}{2} R_{3} \\ &R_{2} \rightarrow R_{2}+3 R_{3} \\ &\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -2 & \frac{1}{2} & 1 \\ 11 & -1 & -6 \\ 4 & \frac{-1}{2} & -2 \end{array}\right] A \end{aligned}$
So, $A^{-1}=\left[\begin{array}{ccc} -2 & \frac{1}{2} & 1 \\ 11 & -1 & -6 \\ 4 & \frac{-1}{2} & -2 \end{array}\right]$

Adjoint and Inverse of Matrix excercise 6.2 question 14

Answer: $\left[\begin{array}{ccc} 3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9 \end{array}\right]$
Hint: Here, we use the concept of matrix inverse using elementary row operation
Given: $\left[\begin{array}{ccc} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}\right]$
Solution: Let $A=\left[\begin{array}{ccc} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}\right]$
$\begin{aligned} &A=I A \\ &A=\left[\begin{array}{ccc} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}\right], I=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{lll} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A \end{aligned}$
Applying $R_{1} \rightarrow \frac{1}{3} R_{1}$
$\Rightarrow\left[\begin{array}{lll} 1 & 0 & \frac{-1}{3} \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}\right]=\left[\begin{array}{lll} \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying $R_{2} \rightarrow R_{2}-2 R_{1}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & \frac{-1}{3} \\ 0 & 3 & \frac{2}{3} \\ 0 & 4 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{-2}{3} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying $R_{2} \rightarrow \frac{1}{3} R_{2}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & \frac{-1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 4 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{-2}{9} & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying $R_{3} \rightarrow R_{3}-4 R_{2}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & \frac{-1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & \frac{1}{9} \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{-2}{9} & \frac{1}{3} & 0 \\ \frac{8}{9} & \frac{-4}{3} & 1 \end{array}\right] A$
Applying $R_{3} \rightarrow 9 R_{3}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & \frac{-1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{-2}{9} & \frac{1}{3} & 0 \\ 8 & -12 & 9 \end{array}\right] A$
Applying
$\begin{aligned} &R_{1} \rightarrow R_{1}+\frac{1}{3} R_{3} \\ &R_{2} \rightarrow R_{2}-\frac{2}{9} R_{3} \end{aligned}$
$\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9 \end{array}\right] A$
So,$A^{-1}=\left[\begin{array}{ccc} 3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9 \end{array}\right]$

Adjoint and inverse of matrix exercise 6.2 question 15

Answer: $A^{-1}=\left[\begin{array}{ccc} 1 & -2 & -3 \\ -2 & 4 & 7 \\ 3 & 5 & 9 \end{array}\right]$
Hint :Here we use basic inverse elementary operation
Given :$\left[\begin{array}{ccc} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{array}\right]$
Solution: $A = IA$
$\left[\begin{array}{ccc} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying $\mathrm{R}_{2} \Rightarrow \mathrm{R}_{2}+3 \mathrm{R}_{{1}}$ and $R_{3} \Rightarrow R_{3}-2 R_{1}$ we get
$\left[\begin{array}{ccc} 1 & 3 & -2 \\ 0 & 9 & -7 \\ 0 & -5 & 4 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{array}\right] A$
Applying $\mathrm{R}_{2} \Rightarrow 1 / 9 \mathrm{R}_{2}$
$\left[\begin{array}{ccc} 1 & 3 & -2 \\ 0 & 1 & -7 / 9 \\ 0 & -5 & 4 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 / 3 & 1 / 9 & 0 \\ -2 & 0 & 1 \end{array}\right] A$
Applying $\mathrm{R}_{{3}} \Rightarrow 9 \mathrm{R}_{3}$
$\left[\begin{array}{ccc} 1 & 0 & 1 / 3 \\ 0 & 1 & -7 / 9 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 0 & -1 / 3 & 0 \\ 1 / 3 & 1 / 9 & 0 \\ -3 & 5 & 9 \end{array}\right] A$
Applying $\mathrm{R}_{1} \Rightarrow \mathrm{R}_{1}-1 / 3 \mathrm{R}_{3}$ and $\mathrm{R}_{2} \Rightarrow \mathrm{R}_{2}+7 / 9 \mathrm{R}_{3}$
$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{array}\right] A$
$I=\left[\begin{array}{ccc} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{array}\right] A$
So $A^{-1}=\left[\begin{array}{ccc} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{array}\right]$

Adjoint and Inverse of Matrix excercise 6.2 question 16

Answer: $\left[\begin{array}{ccc} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{array}\right]$
Hint :Here we use basic inverse elementary operation
Given : $\left[\begin{array}{ccc} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]$
Solution: Let $A=\left[\begin{array}{ccc} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]$
For applying elementary row operation we write,
$\begin{aligned} &A=I A \\ &:\left[\begin{array}{ccc} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A \end{aligned}$
Applying $\mathrm{R}_{1} \leftrightarrow \mathrm{R}_{2}$ we get
$\left[\begin{array}{ccc} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 3 & 1 & 1 \end{array}\right]=\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying $\mathrm{R}_{2} \Rightarrow \mathrm{R}_{2}+\mathrm{R}_{1}$ and $\mathrm{R}_{3} \Rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{1}$ we get
$\Rightarrow\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & -5 & -8 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & -3 & 1 \end{array}\right] A$
Applying $\mathrm{R}_{1} \Rightarrow \mathrm{R}_{1}-2 / 3 \mathrm{R}_{2}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -1 / 3 \\ 0 & 3 & 5 \\ 0 & -5 & -8 \end{array}\right]=\left[\begin{array}{ccc} -2 / 3 & 1 / 3 & 0 \\ 1 & 1 & 0 \\ 0 & -3 & 1 \end{array}\right] A$
Applying $\mathrm{R}_{2} \Rightarrow 1 / 3 \mathrm{R}_{2}$
$\left[\begin{array}{ccc} 1 & 0 & -1 / 3 \\ 0 & 1 & 5 / 3 \\ 0 & -5 & -8 \end{array}\right]=\left[\begin{array}{ccc} -2 / 3 & 1 / 3 & 0 \\ 1 / 3 & 1 / 3 & 0 \\ 0 & -3 & 1 \end{array}\right] A$
Applying $\mathrm{R}_{3} \Rightarrow \mathrm{R}_{3}+5 \mathrm{R}_{2}$
$\left[\begin{array}{ccc} 1 & 0 & -1 / 3 \\ 0 & 1 & 5 / 3 \\ 0 & 0 & 1 / 3 \end{array}\right]=\left[\begin{array}{ccc} -2 / 3 & 1 / 3 & 0 \\ 1 / 3 & 1 / 3 & 0 \\ 5 / 3 & -4 / 3 & 1 \end{array}\right] A$
Applying $\mathrm{R}_{3} \Rightarrow 3 \mathrm{R}_{3}$
$\left[\begin{array}{ccc} 1 & 0 & -1 / 3 \\ 0 & 1 & 5 / 3 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -2 / 3 & 1 / 3 & 0 \\ 1 / 3 & 1 / 3 & 0 \\ 5 & -4 & 3 \end{array}\right] A$
Applying$\mathrm{R}_{1} \Rightarrow \mathrm{R}_{1}+1 / 3 \mathrm{R}_{3}$ and $\mathrm{R}_{2} \Rightarrow \mathrm{R}_{2}-5 / 3 \mathrm{R}_{3}$
$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{array}\right] A$
Hence
$A^{-1}=\left[\begin{array}{ccc} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{array}\right]$

Adjoint and Inverse of Matrix excercise 6.2 question 17

Answer: $\left[\begin{array}{ccc} 3 & -2 & 1 \\ -4 & 1 & -2 \\ 2 & 0 & 1 \end{array}\right]$
Hint: Here, we use the concept of matrix inverse using elementary row operation
Given: $A=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{array}\right]$
Solution: $A=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{array}\right]$
$\begin{aligned} &A=I A \\ &A=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{array}\right], I=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying $R_{2} \rightarrow R_{2}+R_{3}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 2 \\ -2 & -4 & -5 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right] A$
Applying $R_{3} \rightarrow 2 R_{1}+R_{3}$
$\Rightarrow\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{array}\right] A$
Applying $R_{1} \rightarrow R_{1}-3 R_{3}$
$\Rightarrow\left[\begin{array}{lll} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -5 & 0 & -3 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{array}\right] A$
Applyinng $R_{2} \rightarrow R_{2}-2 R_{3}$
$\Rightarrow\left[\begin{array}{lll} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -5 & 0 & -3 \\ -4 & 1 & -2 \\ 2 & 0 & 1 \end{array}\right] A$
Applying $R_{1} \rightarrow R_{1}-2 R_{2}$
$\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 3 & -2 & 1 \\ -4 & 1 & -2 \\ 2 & 0 & 1 \end{array}\right] A$
So,$A^{-1}=\left[\begin{array}{ccc} 3 & -2 & 1 \\ -4 & 1 & -2 \\ 2 & 0 & 1 \end{array}\right]$

Adjoint and Inverse of Matrix excercise 6.2 question 18

Answer: $\left[\begin{array}{ccc} 0 & 1 & -3 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right]$
Hint: Here, we use the concept of matrix inverse using elementary row operation
Given: $\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]$
Solution : Let $A=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]$
$\begin{aligned} &A=I A \\ &A=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right], I=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &\Rightarrow\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A \end{aligned}$
Applying $\mathrm{R}_{1} \leftrightarrow \mathrm{R}_{3}$
$\left[\begin{array}{ccc} 1 & 1 & -2 \\ 3 & 2 & -4 \\ 2 & -3 & 5 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right] A$
Applying $\mathrm{R}_{2} \Rightarrow \mathrm{R}_{2}-3 \mathrm{R}_{1}$ and $\mathrm{R}_{3} \Rightarrow \mathrm{R}_{3}-2 \mathrm{R}_{1}$
$\left[\begin{array}{ccc} 1 & 1 & -2 \\ 0 & -1 & 2 \\ 0 & -5 & 9 \end{array}\right]=\left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & -3 \\ 1 & 0 & -2 \end{array}\right] A$
Applying $\mathrm{R}_{2} \Rightarrow-\mathrm{R}_{2}$ and $\mathrm{R}_{1}=\mathrm{R}_{1+} \mathrm{R}_{2}$
$\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & -5 & 9 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & -3 \\ 0 & -1 & 3 \\ 1 & 0 & -2 \end{array}\right] A$
Applying $\mathrm{R}_{3} \Rightarrow \mathrm{R}_{3}+5 \mathrm{R}_{2}$
$\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & -3 \\ 0 & -1 & 3 \\ 1 & -5 & 13 \end{array}\right] A$
Applying $\mathrm{R}_{3} \Rightarrow-\mathrm{R}_{3}$
$\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & -3 \\ 0 & -1 & 3 \\ -1 & 5 & -13 \end{array}\right] A$
Applying $\mathrm{R}_{2} \Rightarrow \mathrm{R}_{2}+2 \mathrm{R}_{3}$
$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & -3 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right] A$
So, $A^{-1}=\left[\begin{array}{ccc} 0 & 1 & -3 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right]$

RD Sharma Class 12th Exercise 6.2 deals with the chapter Adjoint and Inverse of Matrix. It has 18 Level 1 questions that are relatively easy and can be completed in one go. The questions from this chapter are divided into Level 1 and Level 2, depending on their complexity and weightage.

The level one questions are pretty straightforward and take very little time to complete. Although Level 2 questions for Matrices are also on the more accessible side, they can still be lengthy and time-consuming. However, this particular exercise only has sums.

Chapter-wise RD Sharma Class 12 Solutions