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

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(D) First,calculate the determinant $|A|$: $|A| = 1(3 	imes 3 - 4 	imes 4) - 2(1 	imes 3 - 4 	imes 3) + 3(1 	imes 4 - 3 	imes 3)$ $|A| = 1(9 - 1

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$-\frac{1}{4} \begin{bmatrix} -7 & 6 & -1 \ 9 & -6 & -1 \ -5 & 2 & 1 \end{bmatrix}$

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(D) First,calculate the determinant $|A|$: $|A| = 1(3 	imes 3 - 4 	imes 4) - 2(1 	imes 3 - 4 	imes 3) + 3(1 	imes 4 - 3 	imes 3)$ $|A| = 1(9 - 16) - 2(3 - 12) + 3(4 - 9)$ $|A| = 1(-7) - 2(-9) + 3(-5) = -7 + 18 - 15 = -4$. Next,find the matrix of cofactors $C_{ij}$: $C_{11} = +(9-16) = -7, C_{12} = -(3-12) = 9, C_{13} = +(4-9) = -5$ $C_{21} = -(6-12) = 6, C_{22} = +(3-9) = -6, C_{23} = -(4-6) = 2$ $C_{31} = +(8-9) = -1, C_{32} = -(4-3) = -1, C_{33} = +(3-2) = 1$ Thus,$	ext{adj}(A) = \begin{bmatrix} -7 & 6 & -1 \ 9 & -6 & -1 \ -5 & 2 & 1 \end{bmatrix}$. Finally,$A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{-4} \begin{bmatrix} -7 & 6 & -1 \ 9 & -6 & -1 \ -5 & 2 & 1 \end{bmatrix} = -\frac{1}{4} \begin{bmatrix} -7 & 6 & -1 \ 9 & -6 & -1 \ -5 & 2 & 1 \end{bmatrix}$.

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

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