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(C) Given $A = \begin{bmatrix} 1 & 1 & 2 \ 1 & 3 & 4 \ 1 & -1 & 3 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = 1(3 	imes 3 - 4 	im

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$8$

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(C) Given $A = \begin{bmatrix} 1 & 1 & 2 \ 1 & 3 & 4 \ 1 & -1 & 3 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = 1(3 	imes 3 - 4 	imes -1) - 1(1 	imes 3 - 4 	imes 1) + 2(1 	imes -1 - 3 	imes 1)$$|A| = 1(9 + 4) - 1(3 - 4) + 2(-1 - 3)$$|A| = 13 + 1 - 8 = 6$.We are given $B = \operatorname{adj} A$ and $C = 3A$. We need to find $\frac{|\operatorname{adj} B|}{|C|}$.Since $B = \operatorname{adj} A$,then $\operatorname{adj} B = \operatorname{adj}(\operatorname{adj} A)$.Using the property $|\operatorname{adj} M| = |M|^{n-1}$ where $n$ is the order of the matrix:$|\operatorname{adj} B| = |\operatorname{adj}(\operatorname{adj} A)| = |A|^{(n-1)^2} = |A|^{(3-1)^2} = |A|^4$.For the denominator,$|C| = |3A| = 3^n |A| = 3^3 |A| = 27 |A|$.Thus,$\frac{|\operatorname{adj} B|}{|C|} = \frac{|A|^4}{27 |A|} = \frac{|A|^3}{27}$.Substituting $|A| = 6$:$\frac{6^3}{27} = \frac{216}{27} = 8$.

If the matrices $A = \begin{bmatrix} 1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & -1 & 3 \end{bmatrix}$,$B = \operatorname{adj} A$,and $C = 3A$,then $\frac{|\operatorname{adj} B|}{|C|}$ is equal to

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