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Answer: $A^{-1}=\frac{1}{5}\begin{bmatrix} 1 & 2\\ 2& -1 \end{bmatrix}$

Hint: Here, we use the concept of matrix inverse using elementary row operation

Given: $\begin{bmatrix} 1 &2 \\ 2& -1 \end{bmatrix}$

Solution: Let     $A = \begin{bmatrix} 1 &2 \\ 2& -1 \end{bmatrix}$

$A = IA$

$A = \begin{bmatrix} 1 &2 \\ 2& -1 \end{bmatrix}$,      $I =\begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 1 & 2\\ 2& -1 \end{bmatrix}=\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}A$

Applying $R_{2} \rightarrow R_{2}-2 R_{1}$
$\Rightarrow\left[\begin{array}{cc} 1 & 2 \\ 0 & -5 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ -2 & 1 \end{array}\right] A$

Applying $R_{2} \rightarrow-\frac{1}{5} R_{2}$
$\Rightarrow\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ \frac{2}{5} & \frac{-1}{5} \end{array}\right] A$

Applying    $R_{1} \rightarrow R_{1}-2 R_{2}$
$\Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} \frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & \frac{-1}{5} \end{array}\right] A$

$\Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\frac{1}{5}\left[\begin{array}{cc} 1 & 2 \\ 2 & -1 \end{array}\right] A$

Hence, $A^{-1}=\frac{1}{5}\begin{bmatrix} 1 & 2\\ 2& -1 \end{bmatrix}$

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