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

$\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$

We know

$A=I A$

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

$\Rightarrow\left[\begin{array}{lll}1 & -1 & \frac{4}{3} \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]=\left[\begin{array}{lll}\frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A \quad\left[\right.$ Applying $\left.R_{1} \rightarrow \frac{1}{3} R_{1}\right]$                 [Applying $R_{2} \rightarrow R_{2}-2 R_{1}$ ]

$\Rightarrow\left[\begin{array}{ccc}1 & -1 & \frac{4}{3} \\ 0 & 1 & \frac{-4}{3} \\ 0 & -1 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{3} & 0 & 0 \\ \frac{2}{3} & -1 & 0 \\ 0 & 0 & 1\end{array}\right] A$          [Applying $R_{2} \rightarrow-R_{2}$ ]

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & \frac{-4}{3} \\ 0 & 0 & \frac{-1}{3}\end{array}\right]=\left[\begin{array}{ccc}1 & -1 & 0 \\ \frac{2}{3} & -1 & 0 \\ \frac{2}{3} & -1 & 1\end{array}\right] A$          [Applying $R_{1} \rightarrow R_{1}+R_{2}$ and $R_{3} \rightarrow R_{3}+R_{2}$ ]

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & \frac{-4}{3} \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & -1 & 0 \\ \frac{2}{3} & -1 & 0 \\ -2 & 3 & -3\end{array}\right] A$              $\left[\right.$ Applying $\left.R_{3} \rightarrow-3 R_{3}\right]$

$\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right] A$      $\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}+\frac{4}{3} R_{3}\right]$

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