If A = begin{bmatrix} 1 & -1 & 1 0 & 2 & -3 | vedclass

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(D) First,calculate the determinant of $A$: $|A| = 1(2 	imes 3 - (-3) 	imes 1) - (-1)(0 	imes 3 - (-3) 	imes 2) + 1(0 	imes 1 - 2 	imes 2)$ $|A|

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$1$

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(D) First,calculate the determinant of $A$: $|A| = 1(2 	imes 3 - (-3) 	imes 1) - (-1)(0 	imes 3 - (-3) 	imes 2) + 1(0 	imes 1 - 2 	imes 2)$ $|A| = 1(6 + 3) + 1(0 + 6) + 1(0 - 4) = 9 + 6 - 4 = 11$. Given $B = \operatorname{adj} A$,we know $|B| = |A|^{n-1}$,where $n=3$ is the order of the matrix. $|B| = |A|^{3-1} = |A|^2 = 11^2 = 121$. Now,$|\operatorname{adj} B| = |B|^{n-1} = |B|^2 = 121^2 = 14641$. Given $C = 5A$,then $|C| = |5A| = 5^3 |A| = 125 	imes 11 = 1375$. Finally,$\frac{|\operatorname{adj} B|}{|C|} = \frac{14641}{1375} = 10.647$. Note: Based on the provided options,if the matrix $A$ was intended to result in $|A|=5$,the result would be $1$. Given the matrix provided,the calculation is as shown.

If $A = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 3 \end{bmatrix}$ and $B = \operatorname{adj} A$,$C = 5A$,then $\frac{|\operatorname{adj} B|}{|C|} = $

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