A) \[\left[ \begin{matrix} 5 & 1 & -3 \\ 3 & 2 & 6 \\ 14 & 5 & 0 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 11 & 4 & 3 \\ 1 & 2 & 3 \\ 0 & 3 & 3 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 1 & 8 & 4 \\ 2 & 9 & 6 \\ 0 & 2 & 0 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 0 & 1 & 2 \\ 5 & 4 & 3 \\ 1 & 8 & 2 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
\[A=\left[ \begin{matrix} 1 & 2 & -1 \\ 3 & 0 & 2 \\ 4 & 5 & 0 \\ \end{matrix} \right]\], \[B=\left[ \begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 1 & 3 \\ \end{matrix} \right]\] \[AB=\left[ \begin{matrix} 1\times 1+2\times 2+(-1)(0) \\ 3\times 1+0\times 2+2\times 0 \\ 4\times 1+5\times 2+0\times 0 \\ \end{matrix} \right.\]\[\left. \begin{matrix} 1\times 0+2\times 1+(-1)\,(1) \\ 3\times 0+0\times 1+2\times 1 \\ 4\times 0+5\times 1+0\times 1 \\ \end{matrix}\begin{matrix} \,\,\,\,1\times 0+2\times 0+(-1)\,(3) \\ 3\times 0+0\times 0+2\times 3 \\ 4\times 0+5\times 0+0\times 3 \\ \end{matrix} \right]\] \[\therefore \,\,AB=\left[ \begin{matrix} 5 & 1 & -3 \\ 3 & 2 & 6 \\ 14 & 5 & 0 \\ \end{matrix} \right]\].
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