By using elementary transformation find the inverce of $A = [\begin{matrix} 12 \end{matrix} \begin{matrix} 2 - 1 \end{matrix}]$ .

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Solution

Solution:- Given $A = [\begin{matrix} 12 \end{matrix} \begin{matrix} 2 - 1 \end{matrix}]$
Now we find $A^{- 1}$ by elementary row transformation
so, we can write
$A = A . I$
$[\begin{matrix} 12 \end{matrix} \begin{matrix} 2 - 1 \end{matrix}] = [\begin{matrix} 10 \end{matrix} \begin{matrix} 01 \end{matrix}] . A$
$R_{2} \to R_{2} - 2 R_{1}$
$[\begin{matrix} 10 \end{matrix} \begin{matrix} 2 - 5 \end{matrix}] = [\begin{matrix} 1 - 2 \end{matrix} \begin{matrix} 01 \end{matrix}] . A$
$R_{1} \to R_{1} + \frac{2}{5} \times R_{2}$
$[\begin{matrix} 10 \end{matrix} \begin{matrix} 0 - 5 \end{matrix}] = ⎡ ⎣ \begin{matrix} 1 / 5 - 2 \end{matrix} \begin{matrix} \frac{2}{5} 1 \end{matrix} ⎤ ⎦ . A$
$R_{2} \to R_{2} / - 5$
$[\begin{matrix} 10 \end{matrix} \begin{matrix} 01 \end{matrix}] = ⎡ ⎢ ⎣ \begin{matrix} 1 / 5 \frac{2}{5} \end{matrix} \begin{matrix} 2 / 5 - 1 / 5 \end{matrix} ⎤ ⎥ ⎦ . A$
this from is $I = A^{- 1} . A$ So, $A^{- 1} = [\begin{matrix} 1 / 5 2 / 5 \end{matrix} \begin{matrix} 2 / 5 - 1 / 5 \end{matrix}]$

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