How to find the inverse of following matrix in $\mathbb Z/7 \mathbb Z$?

Question

How to find the inverse of following matrix in $\mathbb Z/7 \mathbb Z$?

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

Need some hint-

Answer 1

Here are the details of the computation: \begin{align*} &\begin{bmatrix} \begin{array}{@{}ccc|ccc@{}} 1 & 2 & 0 & 1 & 0 & 0\\0 & 3 & -1 & 0 & 1 & 0\\-2 & 0 & 2 & 0 & 0 & 1 \end{array} \end{bmatrix} \xrightarrow[R_2\leftarrow -2R_2]{R_3\leftarrow R_3 + 2R_1} \begin{bmatrix} \begin{array}{@{}ccc|ccc@{}} 1 & 2 & 0 & 1 & 0 & 0\\0 & 1 & 2 & 0 & -2 & 0\\0 & 4 & 2 & 2 & 0 & 1 \end{array} \end{bmatrix}\\[1ex]&\xrightarrow{R_3\leftarrow R_3 + 3R_2} \begin{bmatrix} \begin{array}{@{}ccc|ccc@{}} 1 & 2 & 0 & 1 & 0 & 0\\0 & 1 & 2 & 0 & -2 & 0\\0 & 0 & 1 & 2 & 1 & 1 \end{array} \end{bmatrix}\xrightarrow{R_2\leftarrow R_2 - 2R_3}\\[1ex] & \begin{bmatrix} \begin{array}{@{}ccc|ccc@{}} 1 & 2 & 0 & 1 & 0 & 0\\0 & 1 & 0 & 3 &3 & -2 \\0 & 0 & 1 & 2 & 1 & 1 \end{array} \end{bmatrix} \xrightarrow{R_1\leftarrow R_1 - 2R_2} \begin{bmatrix} \begin{array}{@{}ccc|ccc@{}} 1 & 0 & 0 & \color{red}2 & \color{red}1 & \color{red}{-3}\\0 & 1 & 0 & \color{red}3 &\color{red}3 & \color{red}{-2} \\0 & 0 & 1 & \color{red}2 & \color{red}1 & \color{red}1 \end{array} \end{bmatrix} \end{align*}