Let \[A+2B=\left[ \begin{matrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \\ \end{matrix} \right]\] and\[2A-B=\left[ \begin{matrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \\ \end{matrix} \right]\], then \[\operatorname{tr}(A) tr(B)\] is

Let \[A+2B=\left[ \begin{matrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3

1.	Let \[A+2B=\left[ \begin{matrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \\ \end{matrix} \right]\] and\[2A-B=\left[ \begin{matrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \\ \end{matrix} \right]\], then \[\operatorname{tr}(A) tr(B)\] is
A.	1
B.	3
C.	2
D.	0
Answer» D. 0

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Let \[A+2B=\left[ \begin{matrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \\ \end{matrix} \right]\] and\[2A-B=\left[ \begin{matrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \\ \end{matrix} \right]\], then \[\operatorname{tr}(A) tr(B)\] is

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