If A = begin{bmatrix} 3 & -1 -4 & 2 end{bmat | vedclass

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(D) Given $A = \begin{bmatrix} 3 & -1 \ -4 & 2 \end{bmatrix}$.For a $2 	imes 2$ matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$,the invers

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$\begin{bmatrix} 1 & \frac{1}{2} \ 2 & \frac{3}{2} \end{bmatrix}$

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(D) Given $A = \begin{bmatrix} 3 & -1 \ -4 & 2 \end{bmatrix}$.For a $2 	imes 2$ matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$,the inverse is given by $A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$,where $|A| = ad - bc$.First,calculate the determinant $|A| = (3)(2) - (-1)(-4) = 6 - 4 = 2$.Since $|A| 
eq 0$,the inverse exists.Now,find the adjoint matrix: $	ext{adj}(A) = \begin{bmatrix} 2 & 1 \ 4 & 3 \end{bmatrix}$.Therefore,$A^{-1} = \frac{1}{2} \begin{bmatrix} 2 & 1 \ 4 & 3 \end{bmatrix} = \begin{bmatrix} \frac{2}{2} & \frac{1}{2} \ \frac{4}{2} & \frac{3}{2} \end{bmatrix} = \begin{bmatrix} 1 & \frac{1}{2} \ 2 & \frac{3}{2} \end{bmatrix}$.

If $A = \begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}$,then $A^{-1}$ is

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