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

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(C) Step $1$: Calculate the determinant of $A$.$|A| = 1(0 - (-3)) - (-1)(0 - (-6)) + 1(0 - 4) = 1(3) + 1(-6) + 1(-4) = 3 - 6 - 4 = -7$.Step $2$: Use t

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(C) Step $1$: Calculate the determinant of $A$.$|A| = 1(0 - (-3)) - (-1)(0 - (-6)) + 1(0 - 4) = 1(3) + 1(-6) + 1(-4) = 3 - 6 - 4 = -7$.Step $2$: Use the property $|\operatorname{adj} B| = |B|^{n-1}$, where $n=3$.Since $B = \operatorname{adj} A$, we have $|B| = |A|^{n-1} = |A|^2$.Therefore, $|\operatorname{adj} B| = (|A|^2)^{3-1} = (|A|^2)^2 = |A|^4$.Step $3$: Calculate the required ratio $\frac{|\operatorname{adj} B|}{|A|^4}$.$\frac{|\operatorname{adj} B|}{|A|^4} = \frac{|A|^4}{|A|^4} = 1$.

If $A = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0 \end{bmatrix}$, $B = \operatorname{adj} A$ and $C = 5A$, then find the value of $\frac{|\operatorname{adj} B|}{|A|^4}$.

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