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:

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

Solution:

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

To find inverse, first write $A=I A$.

i.e., $\left[\begin{array}{ccc}-1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$

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

$\Rightarrow\left[\begin{array}{ccc}1 & -1 & -2 \\ 0 & 3 & 5 \\ 0 & 4 & 7\end{array}\right]=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 1 & 1 & 0 \\ 3 & 0 & 1\end{array}\right] A$          [Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-3 R_{1}$ ]

$\Rightarrow\left[\begin{array}{ccc}1 & -1 & -2 \\ 0 & 1 & \frac{5}{3} \\ 0 & 4 & 7\end{array}\right]=\left[\begin{array}{ccc}-1 & 0 & 0 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ 3 & 0 & 1\end{array}\right] A$    [Applying $R_{2} \rightarrow \frac{1}{3} R_{2}$ ]

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & \frac{-1}{3} \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & \frac{1}{3}\end{array}\right]=\left[\begin{array}{ccc}\frac{-2}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ \frac{5}{3} & -\frac{4}{3} & 1\end{array}\right] A$    [Applying $R_{3} \rightarrow R_{3}-4 R_{2}$ and $R_{1} \rightarrow R_{1}+R_{2}$ ]

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

$\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3\end{array}\right] A$        [Applying $R_{2} \rightarrow R_{2}-\frac{5}{3} R_{3}$ and $R_{1} \rightarrow R_{1}+\frac{1}{3} R_{3}$ ]

Hence, $A^{-1}=\left[\begin{array}{ccc}1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3\end{array}\right]$