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(A) Step $1$: Calculate the determinant of matrix $A = \begin{bmatrix} 1 & 1 & 2 \ 1 & 2 & 3 \ 2 & 3 & 3 \end{bmatrix}$.$|A| = 1(6-9) - 1(3-6) + 2(3

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$a$

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(A) Step $1$: Calculate the determinant of matrix $A = \begin{bmatrix} 1 & 1 & 2 \ 1 & 2 & 3 \ 2 & 3 & 3 \end{bmatrix}$.$|A| = 1(6-9) - 1(3-6) + 2(3-4) = 1(-3) - 1(-3) + 2(-1) = -3 + 3 - 2 = -2$.Since $a = |adj(A)| = |A|^{n-1}$ where $n=3$,$a = (-2)^{3-1} = (-2)^2 = 4$.Step $2$: Calculate the determinant of matrix $B = \begin{bmatrix} 1 & 2 & 3 \ 4 & -3 & -1 \ 2 & 1 & -4 \end{bmatrix}$.$|B| = 1(12 - (-1)) - 2(-16 - (-2)) + 3(4 - (-6)) = 1(13) - 2(-14) + 3(10) = 13 + 28 + 30 = 71$.Since $b = |B^{-1}| = \frac{1}{|B|} = \frac{1}{71}$.Step $3$: Evaluate $\frac{b+1}{18b}$.$\frac{\frac{1}{71} + 1}{18 	imes \frac{1}{71}} = \frac{\frac{72}{71}}{\frac{18}{71}} = \frac{72}{18} = 4$.Since $a = 4$,the result is $a$.

If $a$ is the determinant of the adjoint of the matrix $A = \begin{bmatrix} 1 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 3 \end{bmatrix}$ and $b$ is the determinant of the inverse of the matrix $B = \begin{bmatrix} 1 & 2 & 3 \\ 4 & -3 & -1 \\ 2 & 1 & -4 \end{bmatrix}$,then find the value of $\frac{b+1}{18b}$.

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