A) \[\left[ \begin{matrix} 7 & 14 \\ 0 & 7 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 5 & 10 \\ 0 & -3 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} -5 & -10 \\ 0 & 3 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 1 & 2 \\ 0 & 2 \\ \end{matrix} \right]\]
Correct Answer: B
Solution :
\[A=\left[ \begin{matrix} 2 & 4 \\ 0 & 3 \\ \end{matrix} \right]\]and\[B=\left[ \begin{matrix} 1 & 2 \\ 0 & 5 \\ \end{matrix} \right]\] \[4A-3B=4\left[ \begin{matrix} 2 & 4 \\ 0 & 3 \\ \end{matrix} \right]-3\left[ \begin{matrix} 1 & 2 \\ 0 & 5 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 8 & 16 \\ 0 & 12 \\ \end{matrix} \right]-\left[ \begin{matrix} 3 & 6 \\ 0 & 15 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 5 & 10 \\ 0 & -3 \\ \end{matrix} \right]\]
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