#### Need solution for RD Sharma maths class 12 chapter 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 $A^{-1}=\left[\begin{array}{ccc} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{array}\right]$