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

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(D) First,calculate the determinant of $A$: $|A| = 1(0 - (-3)) - (-1)(0 - (-6)) + 1(0 - 4) = 1(3) + 1(-6) + 1(-4) = 3 - 6 - 4 = -7$. Note: The propert

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(D) First,calculate the determinant of $A$: $|A| = 1(0 - (-3)) - (-1)(0 - (-6)) + 1(0 - 4) = 1(3) + 1(-6) + 1(-4) = 3 - 6 - 4 = -7$. Note: The property $|	ext{adj}(A)| = |A|^{n-1}$,where $n$ is the order of the matrix. Here $n=3$,so $|	ext{adj}(A)| = |A|^{3-1} = |A|^2$. Given $B = 	ext{adj}(A)$,then $|B| = |A|^2$. We need to find $\frac{|	ext{adj}(B)|}{|C|}$. $|	ext{adj}(B)| = |B|^{n-1} = |B|^2 = (|A|^2)^2 = |A|^4$. Given $C = 5A$,since $A$ is a $3 	imes 3$ matrix,$|C| = |5A| = 5^3 |A| = 125 |A|$. Thus,$\frac{|	ext{adj}(B)|}{|C|} = \frac{|A|^4}{125 |A|} = \frac{|A|^3}{125}$. Wait,re-evaluating the provided solution logic: If $B = 	ext{adj}(A)$,then $	ext{adj}(B) = 	ext{adj}(	ext{adj}(A)) = |A|^{n-2} A = |A|^{3-2} A = |A|A$. So,$|	ext{adj}(B)| = ||A|A| = |A|^3 |A| = |A|^4$. Given $C = 5A$,$|C| = 125|A|$. Ratio = $\frac{|A|^4}{125|A|} = \frac{|A|^3}{125}$. Given the options,let us re-check $|A|$: $|A| = 1(0+3) + 1(0+6) + 1(0-4) = 3 + 6 - 4 = 5$. Then $|	ext{adj}(B)| = |A|^4 = 5^4 = 625$. $|C| = |5A| = 5^3 |A| = 125 	imes 5 = 625$. Therefore,$\frac{|	ext{adj}(B)|}{|C|} = \frac{625}{625} = 1$.

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

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