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$A$ and $B$ are two $3 \times 3$ non-singular matrices such that $\operatorname{adj} A=|A| B$. If $\operatorname{tr}(x)$ denotes the trace of a square matrix $X$ and $C=\left[\begin{array}{ccc}4 & 4 & 7 \\ 3 & -2 & 5 \\ -2 & 3 & 6\end{array}\right]$, then $\sum_{k=1}^x \operatorname{tr}\left(\frac{1}{3^k}(A B)^k C\right)$ is equal to
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$\begin{aligned} & \text { Given, } \operatorname{adj} A=|A| B \\ & \Rightarrow \quad B=\frac{\operatorname{adj} A}{|A|}=A^{-1} \\ & \text { and } \quad \operatorname{tr}(C)=4-2+6=8 \\ & \text { and } \quad(A B)^k=\left(A \cdot A^{-1}\right)^k=(I)^k=I \\ & \text { Now, } \sum_{k=1}^{\infty} \operatorname{tr}\left(\frac{1}{3^k}(A B)^k \cdot C\right)=\sum_{k=1}^{\infty} \operatorname{tr}\left(\frac{1}{3^k} \cdot I \cdot C\right) \\ & =\sum_{k=1}^{\infty} \operatorname{tr}\left(\frac{1}{3^k} \cdot C\right)=\operatorname{tr}(C) \cdot \sum_{k=1}^{\infty} \frac{1}{3^k}=8\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \infty\right] \\ & =8\left[\frac{\frac{1}{3}}{1-\frac{1}{3}}\right]=8\left[\frac{\frac{1}{3}}{\frac{3}{3}}\right] \\ & =8 \times \frac{1}{3} \times \frac{3}{2}=4 \\ & \end{aligned}$
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