Question
Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix}5 & 2 \\ 2 & 1 \end{bmatrix}$

Answer

$\text{A}=\begin{bmatrix}5 & 2 \\ 2 & 1 \end{bmatrix}$For row - transformation A = IA
$\begin{bmatrix}5 & 2 \\ 2 & 1 \end{bmatrix}=\begin{bmatrix} & 0 \\ 0 & 1 \end{bmatrix}\text{A}$
$\text{Applying R}_1\rightarrow\ \frac{1}{5}\text{ R}_1$
$\begin{bmatrix} 1 & \frac{2}{5} \\ 2 & 1 \end{bmatrix}=\begin{bmatrix}\frac{1}{5} & 0 \\ 0 & 1 \end{bmatrix}\text{A}$
Applying $R_1 → R_2 - 2R_1$
$\begin{bmatrix} 1 & \frac{2}{5} \\ 0 & \frac{1}{5} \end{bmatrix}=\begin{bmatrix} \frac{1}{5} & 0 \\ -\frac{2}{5} & 1 \end{bmatrix}\text{A}$
Applying $R_2 → 5R_2​​​​​​​$
$\begin{bmatrix}1 & \frac{2}{5} \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} \frac{1}{5} & 0 \\ -2 & 5 \end{bmatrix}\text{A}$
$\text{Applying R}_1\rightarrow\ \text{R}_1-\frac{2}{5}\text{R}_2$
$\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix}5 & -2 \\ -2 & 1 \end{bmatrix}\text{A}$
Hence, $\text{B}=\begin{bmatrix}5 & -2 \\ -2 & 1 \end{bmatrix}$ is the inverse of A.

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