Using elementary transformations, find the inverse of the followng matrix.

$[\begin{matrix} 1 & - 1 2 & 3 \end{matrix}]$

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Solution

Let $A = [\begin{matrix} 1 & - 1 2 & 3 \end{matrix}]$ ,We know that A=IA
$∴ [\begin{matrix} 1 & - 1 2 & 3 \end{matrix}] = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & - 1 0 & 5 \end{matrix}] = [\begin{matrix} 1 & 0 - 2 & 1 \end{matrix}] A$ (Using $R_{2} \to R_{2} - 2 R_{1})$
$\Rightarrow [\begin{matrix} 1 & - 1 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 \frac{- 2}{5} & \frac{1}{5} \end{matrix}] A$ (Using $R_{2} \to \frac{1}{5} R_{2})$
$\Rightarrow [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] = ⎡ ⎣ \begin{matrix} \frac{3}{5} & \frac{1}{5} \frac{- 2}{5} & \frac{1}{5} \end{matrix} ⎤ ⎦ A$ (Using $R_{1} \to R_{1} + R_{2})$
$∴ A^{- 1} = ⎡ ⎣ \begin{matrix} \frac{3}{5} & \frac{1}{5} \frac{- 2}{5} & \frac{1}{5} \end{matrix} ⎤ ⎦ (∵ A A^{- 1} = I)$

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