By using the elementary transformation, find the inverse of the matrix $A = [\begin{matrix} 1 & - 2 2 & 1 \end{matrix}]$ .

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Solution

Given, $A = [\begin{matrix} 1 & - 2 2 & 1 \end{matrix}]$

Now, $A = I A$

$[\begin{matrix} 1 & - 2 2 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] A$

$R_{1} \to R_{1} + 2 R_{2}$

$\Rightarrow [\begin{matrix} 5 & 0 2 & 1 \end{matrix}] = [\begin{matrix} 1 & 2 0 & 1 \end{matrix}] A$

$R_{1} \to R_{1} / 5$

$\Rightarrow [\begin{matrix} 1 & 0 2 & 1 \end{matrix}] = [\begin{matrix} 1 / 5 & 2 / 5 0 & 1 \end{matrix}] A$

$R_{2} \to R_{2} - 2 R_{1}$

$\Rightarrow A^{- 1} = [\begin{matrix} 1 / 5 & 2 / 5 - 2 / 5 & 1 / 5 \end{matrix}]$

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