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

Question:

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

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

Solution:

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

We know

$A=I A$

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

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

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

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & -\frac{1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 4 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{3} & 0 & 0 \\ -\frac{2}{9} & \frac{1}{3} & 0 \\ 0 & 0 & 1\end{array}\right] A$      [Applying $R_{2} \rightarrow \frac{1}{3} R_{2}$ ]

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & -\frac{1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & \frac{1}{9}\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{3} & 0 & 0 \\ -\frac{2}{9} & \frac{1}{3} & 0 \\ \frac{8}{9} & \frac{-4}{3} & 1\end{array}\right] A$          [Applying $R_{3} \rightarrow R_{3}-4 R_{2}$ ]

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & -\frac{1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{3} & 0 & 0 \\ -\frac{2}{9} & \frac{1}{3} & 0 \\ 8 & -12 & 9\end{array}\right] A$                [Applying $R_{3} \rightarrow 9 R_{3}$ ]

$\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9\end{array}\right] A$    [Applying $R_{2} \rightarrow R_{2}-\frac{2}{9} R_{3}$ and $R_{1} \rightarrow R_{1}+\frac{1}{3} R_{3}$ ]

$A^{-1}=\left[\begin{array}{ccc}3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9\end{array}\right]$