Using elementary tansormations, find the inverse of each of the matrices, if it exists in
$[\begin{matrix} 2 & 1 7 & 4 \end{matrix}]$

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Solution

$A = [\begin{matrix} 2 & 1 7 & 4 \end{matrix}]$

$| A | = 8 - 7 = 1 \neq 0$
Hence, $A^{- 1}$ exists.
Now, $A = I A$
$[\begin{matrix} 2 & 1 7 & 4 \end{matrix}] = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] A$

$⎡ ⎣ \begin{matrix} 1 & \frac{1}{2} 7 & 4 \end{matrix} ⎤ ⎦ = ⎡ ⎣ \begin{matrix} \frac{1}{2} & 0 0 & 1 \end{matrix} ⎤ ⎦ A$ $(R_{1} \to \frac{1}{2} R_{1})$

$⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & \frac{1}{2} 0 & \frac{1}{2} \end{matrix} ⎤ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{2} & 0 - \frac{7}{2} & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦ A$ $(R_{2} \to R_{2} - 7 R_{1})$

$⎡ ⎣ \begin{matrix} 1 & \frac{1}{2} 0 & 1 \end{matrix} ⎤ ⎦ = ⎡ ⎣ \begin{matrix} \frac{1}{2} & 0 - 7 & 2 \end{matrix} ⎤ ⎦ A$ $(R_{2} \to 2 R_{2})$

$[\begin{matrix} 1 & 0 0 & 1 \end{matrix}] = [\begin{matrix} 4 & - 1 - 7 & 2 \end{matrix}] A$ $(R_{1} \to R_{1} - \frac{1}{2} R_{2})$

Hence, $A^{- 1} = [\begin{matrix} 4 & - 1 - 7 & 2 \end{matrix}]$

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