Matrices — Class 12 Maths Solution

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Let us take A =$\left[ {\begin{array}{cccccccccccccccccccc}2&{ - 3}\\{ - 1}&2\end{array}} \right]$

We know that, A = IA
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}2&{ - 3}\\{ - 1}&2\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right]A$

Applying ${R_1} \to {R_1} + {R_2}$
$\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\{ - 1}&2\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&1\\0&1\end{array}} \right]A$

Applying ${R_2} \to {R_2} + {R_1}$
$\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&1\\1&2\end{array}} \right]A$

Applying ${R_1} \to {R_1} + {R_2}$
$\left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}2&3\\1&2\end{array}} \right]A$

Hence, ${A^{ - 1}} = \left[ {\begin{array}{cccccccccccccccccccc}2&3\\1&2\end{array}} \right]$