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

$[\begin{matrix} 3 & - 1 - 4 & 2 \end{matrix}]$

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Solution

let $A = [\begin{matrix} 3 & - 1 - 4 & 2 \end{matrix}]$ . We know that A=IA
$∴ [\begin{matrix} 3 & - 1 - 4 & 2 \end{matrix}] = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & - \frac{1}{3} - 4 & 2 \end{matrix}] = [\begin{matrix} \frac{1}{3} & 0 0 & 1 \end{matrix}] A$ (Using $R_{1} \to \frac{1}{3} R_{1})$
$\Rightarrow ⎡ ⎣ \begin{matrix} 1 & - \frac{1}{3} 0 & \frac{2}{3} \end{matrix} ⎤ ⎦ = ⎡ ⎣ \begin{matrix} \frac{1}{3} & 0 \frac{4}{3} & 1 \end{matrix} ⎤ ⎦ A$ (Using $R_{2} \to R_{2} + 4 R_{1})$
$\Rightarrow [\begin{matrix} 1 & - \frac{1}{3} 0 & 1 \end{matrix}] = ⎡ ⎣ \begin{matrix} \frac{1}{3} & 0 2 & \frac{3}{2} \end{matrix} ⎤ ⎦ A$ (Using $R_{2} \to \frac{3}{2} R_{2})$
$\Rightarrow [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] = [\begin{matrix} 1 & \frac{1}{2} 2 & \frac{3}{2} \end{matrix}] A$ (Using $R_{1} \to R_{1} + \frac{1}{3} R_{2})$
$∴ A^{- 1} = [\begin{matrix} 1 & \frac{1}{2} 2 & \frac{3}{2} \end{matrix}] (∵ A A^{- 1} = I)$

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