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

$[\begin{matrix} 2 & 3 5 & 7 \end{matrix}]$

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Solution

Let $A = [\begin{matrix} 2 & 3 5 & 7 \end{matrix}]$ .We know that A=IA
$[\begin{matrix} 2 & 3 5 & 7 \end{matrix}] = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & \frac{3}{2} 5 & 7 \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 0 & 1 \end{matrix}] A (R_{1} n \to \frac{1}{2} R_{1}) \Rightarrow [\begin{matrix} 1 & \frac{3}{2} 0 & - \frac{1}{2} \end{matrix}] = ⎡ ⎣ \begin{matrix} \frac{1}{2} & 0 \frac{- 5}{2} & 1 \end{matrix} ⎤ ⎦ A$ (Using $R_{2} \to R_{2} - 5 R_{1})$
$\Rightarrow [\begin{matrix} 1 & \frac{3}{2} 0 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 5 & - 2 \end{matrix}] A$ (Using $R_{2} \to (- 2) R_{2})$
$\Rightarrow [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] = [\begin{matrix} - 7 & 3 5 & - 2 \end{matrix}] A$ (Using $R_{1} \to R_{1} - \frac{3}{2} R_{2})$
$A^{- 1} = [\begin{matrix} - 7 & 3 5 & - 2 \end{matrix}]$

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