If A = begin{bmatrix} 2 & 2 9 & 4 end{bmatri | vedclass

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(B) Given $A = \begin{bmatrix} 2 & 2 \ 9 & 4 \end{bmatrix}$.First,calculate the determinant $|A| = (2 	imes 4) - (2 	imes 9) = 8 - 18 = -10$.Next,f

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$A - 6I$

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(B) Given $A = \begin{bmatrix} 2 & 2 \ 9 & 4 \end{bmatrix}$.First,calculate the determinant $|A| = (2 	imes 4) - (2 	imes 9) = 8 - 18 = -10$.Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements: $	ext{adj}(A) = \begin{bmatrix} 4 & -2 \ -9 & 2 \end{bmatrix}$.The inverse is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{-10} \begin{bmatrix} 4 & -2 \ -9 & 2 \end{bmatrix}$.Therefore,$10 A^{-1} = 10 	imes \left( \frac{1}{-10} \begin{bmatrix} 4 & -2 \ -9 & 2 \end{bmatrix} ight) = -1 \begin{bmatrix} 4 & -2 \ -9 & 2 \end{bmatrix} = \begin{bmatrix} -4 & 2 \ 9 & -2 \end{bmatrix}$.Now,check the options:$A - 6I = \begin{bmatrix} 2 & 2 \ 9 & 4 \end{bmatrix} - \begin{bmatrix} 6 & 0 \ 0 & 6 \end{bmatrix} = \begin{bmatrix} 2-6 & 2-0 \ 9-0 & 4-6 \end{bmatrix} = \begin{bmatrix} -4 & 2 \ 9 & -2 \end{bmatrix}$.Thus,$10 A^{-1} = A - 6I$.

If $A = \begin{bmatrix} 2 & 2 \\ 9 & 4 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,then $10 A^{-1}$ is equal to

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