If P and Q are two 3 times 3 matrices such th | vedclass

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(A) Given that $P$ and $Q$ are $3 	imes 3$ matrices,we have $|PQ| = |P||Q| = 1$. Since $|P| = 9$,we get $|Q| = \frac{1}{9}$. We need to find $|	ext{

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$9^4$

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(A) Given that $P$ and $Q$ are $3 	imes 3$ matrices,we have $|PQ| = |P||Q| = 1$. Since $|P| = 9$,we get $|Q| = \frac{1}{9}$. We need to find $|	ext{adj}(P \cdot 	ext{adj}(3Q))|$. Using the property $|	ext{adj}(A)| = |A|^{n-1}$ for an $n 	imes n$ matrix,here $n=3$,so $|	ext{adj}(A)| = |A|^2$. Thus,$|	ext{adj}(P \cdot 	ext{adj}(3Q))| = |P \cdot 	ext{adj}(3Q)|^2 = |P|^2 \cdot |	ext{adj}(3Q)|^2$. Since $|	ext{adj}(3Q)| = |3Q|^{3-1} = |3Q|^2 = (3^3 |Q|)^2 = (27 |Q|)^2$. Substituting $|Q| = \frac{1}{9}$,we get $|	ext{adj}(3Q)| = (27 	imes \frac{1}{9})^2 = 3^2 = 9$. Now,$|	ext{adj}(P \cdot 	ext{adj}(3Q))| = |P|^2 \cdot (9)^2 = 9^2 \cdot 9^2 = 9^4$.

If $P$ and $Q$ are two $3 \times 3$ matrices such that $|PQ|=1$ and $|P|=9$,then the determinant of $\text{adj}(P \cdot \text{adj}(3Q))$ is

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