Using elementary transformations, find the inverse of each of the matrices, if it exists.
$[\begin{matrix} 1 & - 1 2 & 3 \end{matrix}]$

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Solution

$A = I A$ [row transformation]
$[\begin{matrix} 1 & - 1 2 & 3 \end{matrix}] = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] A$
$R_{2} \to R_{2} - 2 R_{1}$
$[\begin{matrix} 1 & - 1 0 & 5 \end{matrix}] = [\begin{matrix} 1 & 0 - 2 & 1 \end{matrix}] A$
$R - 2 \to \frac{1}{5} R_{2}$
$[\begin{matrix} 1 & - 1 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 - 2 / 5 & 1 / 5 \end{matrix}] A$
$R_{1} \to R_{1} + 2 R_{2}$
$[\begin{matrix} 1 & 0 0 & 1 \end{matrix}] = [\begin{matrix} 3 / 5 & 1 / 5 - 2 / 5 & 1 / 5 \end{matrix}] A$
$∴ A^{- 1} = [\begin{matrix} 3 / 5 & 1 / 5 - 2 / 5 & 1 / 5 \end{matrix}]$

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