Let A be a 3 times 3 invertible matrix. If |o | vedclass

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(C) Given that $A$ is a $3 	imes 3$ matrix,so $|A| = \Delta$.Using the property $|\operatorname{adj}(M)| = |M|^{n-1}$ where $n=3$,we have $|\operator

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$2^6$

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(C) Given that $A$ is a $3 	imes 3$ matrix,so $|A| = \Delta$.Using the property $|\operatorname{adj}(M)| = |M|^{n-1}$ where $n=3$,we have $|\operatorname{adj}(M)| = |M|^2$.Given $|\operatorname{adj}(24A)| = |\operatorname{adj}(3 \operatorname{adj}(2A))|$.$|24A|^2 = |3 \operatorname{adj}(2A)|^2$.Since $|kA| = k^n|A|$,$|24A| = 24^3|A|$.So,$(24^3|A|)^2 = (3^3 |\operatorname{adj}(2A)|)^2$.$(24^3|A|)^2 = (27 |2A|^2)^2$.Since $|2A| = 2^3|A| = 8|A|$,we have $|\operatorname{adj}(2A)| = (8|A|)^2 = 64|A|^2$.Substituting this back: $(24^3|A|)^2 = (27 	imes 64|A|^2)^2$.$(24^3|A|)^2 = (1728|A|^2)^2$.Since $24^3 = 13824$,we have $(13824|A|)^2 = (1728|A|^2)^2$.$13824|A| = 1728|A|^2$ (assuming $|A| 
eq 0$).$|A| = \frac{13824}{1728} = 8$.We need to find $|A^2| = |A|^2 = 8^2 = 64 = 2^6$.

Let $A$ be a $3 \times 3$ invertible matrix. If $|\operatorname{adj}(24A)| = |\operatorname{adj}(3 \operatorname{adj}(2A))|$,then $|A^2|$ is equal to

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