#### Please solve rd  sharma class 12 Chapter 6   Adjoint and Inverse of Matrix excercise 6.2 question 16  maths textbook solution

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

$A^{-1}=\left[\begin{array}{ccc} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{array}\right]$