Find the inverse of matrix $A$. Can some please show me how to do this question.I've been attempting this question for quite awhile now, although don't know how to proceed.
$$A=\begin{bmatrix} \frac{1}{5} & \frac{1}{5} & -\frac{2}{5} \\ \frac{1}{5} & \frac{1}{5} & \frac{1}{10}\\ \frac{1}{5} & -\frac{4}{5} & \frac{1}{10} \end{bmatrix} $$
$$\begin{align}\left[\begin{array}{ccc|ccc} \color{blue}{\frac{1}{5}} & \color{blue}{\frac{1}{5}} & \color{blue}{-\frac{2}{5}}& \color{red}{1} & 0 & 0 \\ \color{blue}{\frac{1}{5}} & \color{blue}{\frac{1}{5}} & \color{blue}{\frac{1}{10}} & 0 & \color{red}{1} & 0\\ \color{blue}{\frac{1}{5}} & \color{blue}{-\frac{4}{5}} & \color{blue}{\frac{1}{10}} & 0 & 0 & \color{red}{1} \end{array}\right] &\sim \left[\begin{array}{ccc|ccc} \color{red}{1} & \color{blue}{1} & \color{blue}{-2} & \color{blue}{5} & 0 & 0 \\ \color{blue}{2} & \color{blue}{2} & \color{blue}{1} & 0 & \color{blue}{10} & 0\\ \color{blue}{2} & \color{blue}{-8} & \color{blue}{1} & 0 & 0 & \color{blue}{10} \end{array}\right] & \begin{pmatrix}R_1 \to 5R_1 \\ R_2 \to 10R_2 \\ R_3 \to 10R_3\end{pmatrix}\\ &\sim \left[\begin{array}{ccc|ccc} \color{red}{1} & \color{blue}{1} & \color{blue}{-2} & \color{blue}{5} & 0 & 0 \\ 0 & 0 & \color{blue}{5} & \color{blue}{-10} & \color{blue}{10} & 0\\ 0 & \color{blue}{-10} & \color{blue}{5} & \color{blue}{-10} & 0 & \color{blue}{10} \end{array}\right] & \begin{pmatrix}R_2 \to R_2-2R_1 \\ R_3 \to R_3-2R_1\end{pmatrix} \\ &\sim \left[\begin{array}{ccc|ccc} \color{red}{1} & \color{blue}{1} & \color{blue}{-2} & \color{blue}{5} & 0 & 0 \\ 0 & 0 & \color{blue}{5} & \color{blue}{-10} & \color{blue}{10} & 0\\ 0 & \color{blue}{-10} & 0 & 0 & \color{blue}{-10} & \color{blue}{10} \end{array}\right] & \begin{pmatrix}R_3 \to R_3 - R_2\end{pmatrix} \\ &\sim \left[\begin{array}{ccc|ccc} \color{red}{1} & \color{blue}{1} & \color{blue}{-2} & \color{blue}{5} & 0 & 0 \\ 0 & 0 & \color{red}{1} & \color{blue}{-2} & \color{blue}{2} & 0\\ 0 & \color{red}{1} & 0 & 0 & \color{blue}{1} & \color{blue}{-1} \end{array}\right] & \begin{pmatrix}R_2 \to \frac 15R_2 \\ R_3 \to -\frac 1{10}R_3\end{pmatrix} \\ &\sim \left[\begin{array}{ccc|ccc} \color{red}{1} & 0 & 0 & \color{blue}{1} & \color{blue}{3} & \color{blue}{1} \\ 0 & 0 & \color{red}{1} & \color{blue}{-2} & \color{blue}{2} & 0\\ 0 & \color{red}{1} & 0 & 0 & \color{blue}{1} & \color{blue}{-1} \end{array}\right] & \begin{pmatrix}R_1 \to R_1+2R_2-R_3\end{pmatrix} \\ &\sim \left[\begin{array}{ccc|ccc} \color{red}{1} & 0 & 0 & \color{blue}{1} & \color{blue}{3} & \color{blue}{1} \\ 0 & \color{red}{1} & 0 & 0 & \color{blue}{1} & \color{blue}{-1} \\ 0 & 0 & \color{red}{1} & \color{blue}{-2} & \color{blue}{2} & 0 \end{array}\right] & \begin{pmatrix}R_2 \leftrightarrow R_3\end{pmatrix}\end{align}$$
So the inverse matrix is $$\left[\begin{array}{ccc} 1 & 3 & 1 \\ 0 & 1 & -1 \\ -2 & 2 & 0 \end{array}\right]$$