A and B are two 3 times 3 non-singular matric | vedclass

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(B) Given, $\operatorname{adj} A = |A| B$. Since $A$ is non-singular, $\operatorname{adj} A = |A| A^{-1}$. Thus, $|A| B = |A| A^{-1}$, which implies $

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$4$

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(B) Given, $\operatorname{adj} A = |A| B$. Since $A$ is non-singular, $\operatorname{adj} A = |A| A^{-1}$. Thus, $|A| B = |A| A^{-1}$, which implies $B = A^{-1}$. Therefore, $A B = A A^{-1} = I$, where $I$ is the identity matrix. Now, consider the sum $S = \sum_{k=1}^{\infty} \operatorname{tr}\left(\frac{1}{3^k}(A B)^k Cight)$. Since $(A B)^k = I^k = I$, the expression becomes $S = \sum_{k=1}^{\infty} \operatorname{tr}\left(\frac{1}{3^k} I Cight) = \sum_{k=1}^{\infty} \frac{1}{3^k} \operatorname{tr}(C)$. The trace of $C$ is $\operatorname{tr}(C) = 4 + (-2) + 6 = 8$. So, $S = 8 \sum_{k=1}^{\infty} \left(\frac{1}{3}ight)^k$. This is an infinite geometric series with first term $a = \frac{1}{3}$ and common ratio $r = \frac{1}{3}$. The sum is $S = 8 \left( \frac{1/3}{1 - 1/3} ight) = 8 \left( \frac{1/3}{2/3} ight) = 8 	imes \frac{1}{2} = 4$.

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