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

$[\begin{matrix} 2 & 5 1 & 3 \end{matrix}]$

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Solution

Let $A = [\begin{matrix} 2 & 5 1 & 3 \end{matrix}]$ . We know that A =IA
$∴ [\begin{matrix} 2 & 5 1 & 3 \end{matrix}] = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & 3 2 & 5 \end{matrix}] = [\begin{matrix} 0 & 1 1 & 0 \end{matrix}] A$ (Using $R_{2} \leftrightarrow R_{1})$
$\Rightarrow [\begin{matrix} 1 & 3 0 & - 1 \end{matrix}] = [\begin{matrix} 0 & 1 1 & - 2 \end{matrix}] A (U s i n g R_{2} \to R_{2} - 2 R_{1})$

$\Rightarrow [\begin{matrix} 1 & 3 0 & 1 \end{matrix}] = [\begin{matrix} 0 & 1 - 1 & 2 \end{matrix}] A (U s i n g R_{2} \to (- 1) R_{2}) \Rightarrow [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] = [\begin{matrix} 3 & - 5 - 1 & 2 \end{matrix}] A (U s i n g R_{1} \to R_{1} - 3 R_{2}) ∴ A^{- 1} = [\begin{matrix} 3 & - 5 - 1 & 2 \end{matrix}] (∵ A A^{- 1} = I)$

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