{left[ {begin{array}{*{20}{c}}1&33&{10}end{ar | vedclass

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(B) Let $A = \left[ {\begin{array}{*{20}{c}}1&3\3&{10}\end{array}} ight]$.To find the inverse $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} \

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$\left[ {\begin{array}{*{20}{c}}{10}&{ - 3}\{ - 3}&1\end{array}} ight]$

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(B) Let $A = \left[ {\begin{array}{*{20}{c}}1&3\3&{10}\end{array}} ight]$.To find the inverse $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.First,calculate the determinant $|A| = (1 	imes 10) - (3 	imes 3) = 10 - 9 = 1$.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}}{10}&{-3}\{-3}&1\end{array}} ight]$.Thus,$A^{-1} = \frac{1}{1} \left[ {\begin{array}{*{20}{c}}{10}&{-3}\{-3}&1\end{array}} ight] = \left[ {\begin{array}{*{20}{c}}{10}&{-3}\{-3}&1\end{array}} ight]$.Therefore,the correct option is $B$.

${\left[ {\begin{array}{*{20}{c}}1&3\\3&{10}\end{array}} \right]^{ - 1}} = $

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