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

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(D) First,calculate the determinant $|A|$:$|A| = 1(16 - 9) - 3(4 - 3) + 3(3 - 4) = 1(7) - 3(1) + 3(-1) = 7 - 3 - 3 = 1 
eq 0$.Next,find the cofactor

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$\begin{bmatrix} 7 & -3 & -3 \ -1 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$

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(D) First,calculate the determinant $|A|$:$|A| = 1(16 - 9) - 3(4 - 3) + 3(3 - 4) = 1(7) - 3(1) + 3(-1) = 7 - 3 - 3 = 1 
eq 0$.Next,find the cofactor matrix and its transpose to get $	ext{adj } A$:$A_{11} = +(16-9) = 7, A_{12} = -(4-3) = -1, A_{13} = +(3-4) = -1$$A_{21} = -(12-9) = -3, A_{22} = +(4-3) = 1, A_{23} = -(3-3) = 0$$A_{31} = +(9-12) = -3, A_{32} = -(3-3) = 0, A_{33} = +(4-3) = 1$Thus,$	ext{adj } A = \begin{bmatrix} 7 & -3 & -3 \ -1 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$.Verify $A(	ext{adj } A) = |A|I$:$A(	ext{adj } A) = \begin{bmatrix} 1 & 3 & 3 \ 1 & 4 & 3 \ 1 & 3 & 4 \end{bmatrix} \begin{bmatrix} 7 & -3 & -3 \ -1 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 7-3-3 & -3+3+0 & -3+0+3 \ 7-4-3 & -3+4+0 & -3+0+3 \ 7-3-4 & -3+3+0 & -3+0+4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = 1 \cdot I = |A|I$.Finally,$A^{-1} = \frac{1}{|A|} 	ext{adj } A = \frac{1}{1} \begin{bmatrix} 7 & -3 & -3 \ -1 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -3 & -3 \ -1 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$,then verify that $A \text{ adj } A = |A| I$. Also,find $A^{-1}$.

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