If A = begin{bmatrix} 1 & -5 -2 & 4 end{bmat | vedclass

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(A) Step $1$: Find the determinant of $A$. $|A| = (1)(4) - (-5)(-2) = 4 - 10 = -6$. Step $2$: Find the adjoint of $A$. For a $2 	imes 2$ matrix $\beg

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$-\frac{1}{6} \begin{bmatrix} 4 & 5 \ 2 & 1 \end{bmatrix}$

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(A) Step $1$: Find the determinant of $A$. $|A| = (1)(4) - (-5)(-2) = 4 - 10 = -6$. Step $2$: Find the adjoint of $A$. For a $2 	imes 2$ matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$, the adjoint is $\begin{bmatrix} d & -b \ -c & a \end{bmatrix}$. Step $3$: Thus, $	ext{adj}(A) = \begin{bmatrix} 4 & 5 \ 2 & 1 \end{bmatrix}$. Step $4$: Use the formula $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$. Step $5$: $A^{-1} = \frac{1}{-6} \begin{bmatrix} 4 & 5 \ 2 & 1 \end{bmatrix} = -\frac{1}{6} \begin{bmatrix} 4 & 5 \ 2 & 1 \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & -5 \\ -2 & 4 \end{bmatrix}$, then $A^{-1} =$

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