Let A be a 3 times 3 matrix such that |operat | vedclass

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(A) Given $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))| = 12^4$. We know that for a matrix $A$ of order $n$,$|\operatorname{adj}(\op

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$2 \sqrt{3}$

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(A) Given $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))| = 12^4$. We know that for a matrix $A$ of order $n$,$|\operatorname{adj}(\operatorname{adj}(\dots \operatorname{adj} A))|$ ($k$ times) is $|A|^{(n-1)^k}$. Here $n = 3$ and $k = 3$,so $|A|^{(3-1)^3} = 12^4$. $|A|^{2^3} = 12^4 \Rightarrow |A|^8 = 12^4$. Taking the square root on both sides,$|A|^4 = 12^2 = 144$. $|A|^2 = 12 \Rightarrow |A| = \sqrt{12} = 2\sqrt{3}$. We need to find $|A^{-1} \operatorname{adj} A|$. Using the property $|XY| = |X||Y|$,we have $|A^{-1}| |\operatorname{adj} A|$. Since $|A^{-1}| = \frac{1}{|A|}$ and $|\operatorname{adj} A| = |A|^{n-1} = |A|^{3-1} = |A|^2$. Thus,$|A^{-1} \operatorname{adj} A| = \frac{1}{|A|} \cdot |A|^2 = |A|$. Therefore,$|A^{-1} \operatorname{adj} A| = 2\sqrt{3}$.

Let $A$ be a $3 \times 3$ matrix such that $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A ))|=12^4$. Then $|A^{-1} \operatorname{adj} A|$ is equal to

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