Find the inverse of the following matrices by using elementary row transformations: $\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$

Solution:

For elementary row operation we have,
$\begin{array}{l} A=I A \\ \Rightarrow\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A} \end{array}$
Applying $r_{1} \leftrightarrow r_{2}$
$\Rightarrow\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1 \end{array}\right]=\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}$
Applying $r_{3} \rightarrow r_{3}-3 r_{1}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{array}\right] A$
Applying $r_{1} \rightarrow r_{1}-2 r_{2}$ and $r_{3} \rightarrow r_{3}+5 r_{2}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{array}\right]=\left[\begin{array}{ccc} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1 \end{array}\right] A$
Applying $r_{3} \rightarrow \frac{1}{2} r_{3}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -2 & 1 & 0 \\ 1 & 0 & 0 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{array}\right] A$
Applying $r_{1} \rightarrow r_{1}+r_{3}$ and $r_{2} \rightarrow r_{2}-2 r_{3}$
$\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & 1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array}\right] \mathrm{A}$
Therefore, as it is known that
$I=A^{-1} A$
So
$\Rightarrow^{-1}=\left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & 1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array}\right]$