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

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(D) Given $A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$ and $AX = I$.Since $AX = I$,it follows that $X = A^{-1}$.First,find the determinant of $A

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

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(D) Given $A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$ and $AX = I$.Since $AX = I$,it follows that $X = A^{-1}$.First,find the determinant of $A$: $|A| = (1)(4) - (2)(3) = 4 - 6 = -2$.The inverse of a $2 	imes 2$ matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$ is given by $\frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Therefore,$A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \ -3 & 1 \end{bmatrix}$.$X = \begin{bmatrix} \frac{4}{-2} & \frac{-2}{-2} \ \frac{-3}{-2} & \frac{1}{-2} \end{bmatrix} = \begin{bmatrix} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $X$ is a $2 \times 2$ matrix such that $AX = I$,then $X =$

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