If the inverse of the matrix A = begin{bmatri | vedclass

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(C) First, we find the determinant of $A$: $|A| = -1(5 - 8) - (-3)(0 - 6) + (-2)(0 - 3) = -1(-3) + 3(-6) - 2(-3) = 3 - 18 + 6 = -9$. Next, we find the

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$\frac{2}{3}$

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(C) First, we find the determinant of $A$: $|A| = -1(5 - 8) - (-3)(0 - 6) + (-2)(0 - 3) = -1(-3) + 3(-6) - 2(-3) = 3 - 18 + 6 = -9$. Next, we find the matrix of cofactors $C_{ij}$: $C_{11} = +(5 - 8) = -3$, $C_{12} = -(0 - 6) = 6$, $C_{13} = +(0 - 3) = -3$. $C_{21} = -(-15 - (-8)) = -(-7) = 7$, $C_{22} = +(-5 - (-6)) = 1$, $C_{23} = -(-4 - (-9)) = -5$. $C_{31} = +(-6 - (-2)) = -4$, $C_{32} = -(-2 - 0) = 2$, $C_{33} = +(-1 - 0) = -1$. The adjoint matrix is the transpose of the cofactor matrix: $	ext{adj } A = \begin{bmatrix} -3 & 7 & -4 \ 6 & 1 & 2 \ -3 & -5 & -1 \end{bmatrix}$. Thus, $A^{-1} = \frac{1}{|A|} 	ext{adj } A = -\frac{1}{9} \begin{bmatrix} -3 & 7 & -4 \ 6 & 1 & 2 \ -3 & -5 & -1 \end{bmatrix} = \begin{bmatrix} 1/3 & -7/9 & 4/9 \ -6/9 & -1/9 & -2/9 \ 3/9 & 5/9 & 1/9 \end{bmatrix}$. Comparing this to the given form, we have $a_1 = 1/3$, $c_2 = 5/9$, and $b_3 = -2/9$. Therefore, $a_1 + c_2 + b_3 = \frac{3}{9} + \frac{5}{9} - \frac{2}{9} = \frac{6}{9} = \frac{2}{3}$.

If the inverse of the matrix $A = \begin{bmatrix} -1 & -3 & -2 \\ 0 & 1 & 2 \\ 3 & 4 & 5 \end{bmatrix}$ is $A^{-1} = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$, then find the value of $a_1 + c_2 + b_3$.

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