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

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(A) Given matrix $A = \begin{bmatrix} 1 & 2 \ -4 & -1 \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = (1)(-1) - (2)(-4) = -1 + 8 = 7$

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$\frac{1}{7} \begin{bmatrix} -1 & -2 \ 4 & 1 \end{bmatrix}$

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(A) Given matrix $A = \begin{bmatrix} 1 & 2 \ -4 & -1 \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = (1)(-1) - (2)(-4) = -1 + 8 = 7$.Since $|A| 
eq 0$,$A^{-1}$ exists.The adjoint of a $2 	imes 2$ matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$ is given by $\begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Therefore,$	ext{adj}(A) = \begin{bmatrix} -1 & -2 \ 4 & 1 \end{bmatrix}$.The inverse is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.$A^{-1} = \frac{1}{7} \begin{bmatrix} -1 & -2 \ 4 & 1 \end{bmatrix}$.

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

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