$|A|\,\, = 4 + 6 = 10 \ne 0$
अब, ${A_{11}} = 4$, $a = - 6,\,A = \left| {\,\begin{array}{*{20}{c}}0&0&0\\0&0&0\\1&{ - 2}&{ - 5}\end{array}\,} \right|$, ${A_{21}} = - ( - 2) = 2$, ${A_{22}} = 1$]
$\therefore $ $adj\,(A) = \left[ {\begin{array}{*{20}{c}}4&2\\{ - 3}&1\end{array}} \right]$
$\therefore $ ${A^{ - 1}} = \frac{{adj\,(A)}}{{|A|}} = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}}4&2\\{ - 3}&1\end{array}} \right]$.
ट्रिक : विकल्प के निरीक्षण से, $A{A^{ - 1}} = I\,$
$\Rightarrow$ $A{A^{ - 1}}$ = $\left[ {\begin{array}{*{20}{c}}1&{ - 2}\\3&4\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{\frac{4}{{10}}}&{\frac{2}{{10}}}\\{\frac{{ - 3}}{{10}}}&{\frac{1}{{10}}}\end{array}} \right]$ = $\left[ {\begin{array}{*{20}{c}}{\frac{{10}}{{10}}}&0\\0&{\frac{{10}}{{10}}}\end{array}} \right]$
= $A{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I$.
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