Find the inverse of each of the following matrices by using elementary row transformations:

Solution:

Let $A=\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]$

$A=I A$

$\Rightarrow\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$

Applying $R_{1} \leftrightarrow R_{3}$

$\Rightarrow\left[\begin{array}{ccc}1 & 1 & -2 \\ 3 & 2 & -4 \\ 2 & -3 & 5\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right] A$

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

$\Rightarrow\left[\begin{array}{ccc}1 & 1 & -2 \\ 0 & -1 & 2 \\ 0 & -5 & 9\end{array}\right]=\left[\begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & -3 \\ 1 & 0 & -2\end{array}\right] A$

Applying $R_{2} \rightarrow(-1) R_{2}$

$\Rightarrow\left[\begin{array}{ccc}1 & 1 & -2 \\ 0 & 1 & -2 \\ 0 & -5 & 9\end{array}\right]=\left[\begin{array}{ccc}0 & 0 & 1 \\ 0 & -1 & 3 \\ 1 & 0 & -2\end{array}\right] A$

Applying $R_{3} \rightarrow R_{3}+5 R_{2}$ and $R_{1} \rightarrow R_{1}-R_{2}$

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & -1\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & -2 \\ 0 & -1 & 3 \\ 1 & -5 & 13\end{array}\right] A$

Applying $R_{3} \rightarrow(-1) R_{3}$

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & -2 \\ 0 & -1 & 3 \\ -1 & 5 & -13\end{array}\right] A$

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

$\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right] A$

Hence, $A^{-1}=\left[\begin{array}{ccc}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right]$