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(A) Let $A = \left[ {\begin{array}{*{20}{c}}{ - 6}&5\{ - 7}&6\end{array}} ight]$.To find $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} 	ext{

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$\left[ {\begin{array}{*{20}{c}}{ - 6}&5\{ - 7}&6\end{array}} ight]$

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(A) Let $A = \left[ {\begin{array}{*{20}{c}}{ - 6}&5\{ - 7}&6\end{array}} ight]$.To find $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.First,calculate the determinant $|A| = (-6)(6) - (5)(-7) = -36 + 35 = -1$.Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements: $	ext{adj}(A) = \left[ {\begin{array}{*{20}{c}}6&{-5}\7&{-6}\end{array}} ight]$.Therefore,$A^{-1} = \frac{1}{-1} \left[ {\begin{array}{*{20}{c}}6&{-5}\7&{-6}\end{array}} ight] = \left[ {\begin{array}{*{20}{c}}-6&5\-7&6\end{array}} ight]$.

${\left[ {\begin{array}{*{20}{c}}{ - 6}&5\\{ - 7}&6\end{array}} \right]^{ - 1}}$ =

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