If A is a 3 times 3 matrix and |A|=frac{1}{2} | vedclass

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(A) Given that $A$ is a $3 	imes 3$ matrix,so $n=3$. We know that $\operatorname{Adj}(\operatorname{Adj} A) = |A|^{n-2} A = |A|^{3-2} A = |A| A$. Now

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$8$

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(A) Given that $A$ is a $3 	imes 3$ matrix,so $n=3$. We know that $\operatorname{Adj}(\operatorname{Adj} A) = |A|^{n-2} A = |A|^{3-2} A = |A| A$. Now,consider the expression $|A^{-1}(\operatorname{Adj}(\operatorname{Adj} A))|^{-1}$. Substituting the property: $|A^{-1}(|A| A)|^{-1} = | |A| (A^{-1} A) |^{-1} = | |A| I |^{-1}$. Since $A$ is a $3 	imes 3$ matrix,$| |A| I | = |A|^3 |I| = |A|^3 	imes 1 = |A|^3$. Therefore,the expression becomes $(|A|^3)^{-1} = \frac{1}{|A|^3}$. Given $|A| = \frac{1}{2}$,we have $\frac{1}{(\frac{1}{2})^3} = \frac{1}{\frac{1}{8}} = 8$.

If $A$ is a $3 \times 3$ matrix and $|A|=\frac{1}{2}$,then $|A^{-1}(\operatorname{Adj}(\operatorname{Adj} A))|^{-1} = $

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