#### Provide solution for rd sharma class 12 chapter 6 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,$A^{-1}=\left[\begin{array}{ccc} 3 & -2 & 1 \\ -4 & 1 & -2 \\ 2 & 0 & 1 \end{array}\right]$