Using elementary operations, find the inverse of the following matrix:

$\left(\begin{array}{ccc}- 1 & 1 & 2\\ 1& 2 & 3\\ 3& 1 & 1\end{array} \right)$

Consider the given matrix.

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

We know that,  A = I_n A

Perform sequence of elementary row operations on  A  on the left hand side and the term  I_n  on the right hand side till we obtain the results,

I_n = BA

Thus,  B = A^{- 1} 

$$\begin{gathered} \mathrm{Here}, {\mathrm{I}}_3 = \left[ \begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array} \right] \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}, \\ \left[ \begin{array}{ccc}- 1& 1& 2\\ 1& 2& 3\\ 3& 1& 1\end{array} \right] = \left[ \begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array} \right] \mathrm{A} \\ {\mathrm{R}}_1 \leftrightarrow {\mathrm{R}}_2 \\ \left[\begin{array}{ccc} 1& 2& 3\\ - 1& 1& 2\\ 3& 1& 1\end{array} \right] = \left[ \begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array} \right] \mathrm{A} \\ {\mathrm{R}}_2 \to {\mathrm{R}}_2 + {\mathrm{R}}_1 \\ {\mathrm{R}}_3 \to {\mathrm{R}}_3 - 3 {\mathrm{R}}_1 \\ \left[\begin{array}{ccc} 1 & 2 & 3\\ 0 & 3 & 5\\ 3 & - 5 & - 8 \end{array} \right] = \left[ \begin{array}{ccc}0& 1& 0\\ 1& 1& 0\\ 0& - 3 & 1\end{array} \right] \mathrm{A} \\ {\mathrm{R}}_{1 } \to {\mathrm{R}}_{1 } + {\mathrm{R}}_{2 } \\ \left[\begin{array}{ccc} 1 & 5 & 8\\ 0 & 3 & 5\\ 0 & - 5 & - 8 \end{array} \right] = \left[ \begin{array}{ccc}1& 2& 0\\ 1& 1& 0\\ 0& - 3 & 1\end{array} \right] \mathrm{A} \\ {\mathrm{R}}_{1 } \to {\mathrm{R}}_{1 } + {\mathrm{R}}_3 \end{gathered}$$

$$\begin{gathered} \left[ \begin{array}{ccc}1& 0 & 0 \\ 0& 3 & 5 \\ 0& - 5& - 8 \end{array} \right] = \left[ \begin{array}{ccc}1 & - 1& 1\\ 1 & 1& 0\\ 0 & - 3& 1\end{array} \right] \mathrm{A} \\ {\mathrm{R}}_{2 } \to \frac{{\mathrm{R}}_{2 }}{3} \\ \left[ \begin{array}{ccc}1& 0 & 0 \\ 0& 1 & \frac{5}{3} \\ 0& - 5& - 8 \end{array} \right] = \left[ \begin{array}{ccc}1 & - 1& 1\\ \frac{1}{3} & \frac{1}{3}& 0\\ 0 & - 3 & 1\end{array} \right] \mathrm{A} \\ {\mathrm{R}}_3 \to {\mathrm{R}}_3 + 5 {\mathrm{R}}_2 \\ \left[ \begin{array}{ccc}1& 0 & 0 \\ 0& 1 & \frac{5}{3} \\ 0& 0& \frac{1}{3}\end{array} \right] = \left[ \begin{array}{ccc}1 & - 1& 1\\ \frac{1}{3} & \frac{1}{3}& 0\\ \frac{5}{3} & \frac{- 4}{3}& 1\end{array} \right] \mathrm{A} \\ \left[ \begin{array}{ccc}1& 0 & 0 \\ 0& 1 & \frac{5}{3} \\ 0& 0& 1\end{array} \right] = \left[ \begin{array}{ccc}1 & - 1& 1\\ \frac{1}{3} & \frac{1}{3}& 0\\ 5 & - 4 & 3\end{array} \right] \mathrm{A} \end{gathered}$$

$$\begin{gathered} {\mathrm{R}}_2 \to {\mathrm{R}}_2 - \frac{5}{3} {\mathrm{R}}_3 \\ \left[ \begin{array}{ccc}1& 0& 0\\ 0& 1& 0 \\ 0& 0& 1\end{array} \right] = \left[\begin{array}{ccc} 1 & - 1& 1\\ - 8& 7& - 5\\ 5& - 4& 3\end{array}\right] \mathrm{A} \end{gathered}$$

Thus, the inverse of the matrix  A  is given by 

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