A) Skew-symmetric matrix
B) Symmetric matrix
C) Diagonal matrix
D) None of these
Correct Answer: B
Solution :
[b] Let \[A=\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & 1 & 0 \\ 1 & -1 & 2 \\ \end{matrix} \right],\] then \[A'=\left[ \begin{matrix} 1 & 2 & 1 \\ -1 & 1 & -1 \\ 1 & 0 & 2 \\ \end{matrix} \right]\] \[\therefore AA'=\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & 1 & 0 \\ 1 & -1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 1 \\ -1 & 1 & -1 \\ 1 & 0 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 & 1 & 4 \\ 1 & 5 & 1 \\ 4 & 1 & 4 \\ \end{matrix} \right]\]
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