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#### Provide solution for rd sharma class 12 chapter 6 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, $A^{-1}=\left[\begin{array}{ccc} 0 & 1 & -3 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right]$