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

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(C) Given $AX = I$,then $X = A^{-1}$.First,calculate the determinant of $A$:$|A| = (1 	imes 3) - (2 	imes 4) = 3 - 8 = -5$.Next,find the adjoint of

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$\frac{1}{5} \begin{bmatrix} -3 & 2 \ 4 & -1 \end{bmatrix}$

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(C) Given $AX = I$,then $X = A^{-1}$.First,calculate the determinant of $A$:$|A| = (1 	imes 3) - (2 	imes 4) = 3 - 8 = -5$.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} 3 & -2 \ -4 & 1 \end{bmatrix}$.Now,calculate $X = A^{-1} = \frac{1}{|A|} 	ext{adj } A$:$X = \frac{1}{-5} \begin{bmatrix} 3 & -2 \ -4 & 1 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} -3 & 2 \ 4 & -1 \end{bmatrix}$.

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

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