A) \[\left[ \begin{matrix} 3 & 2 \\ 5 & 3 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 5 & 2 \\ 5 & 5 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 5 & 2 \\ 3 & 5 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 3 & 2 \\ 5 & 5 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
Given,\[A=\left[ \begin{matrix} 2 & -1 \\ 4 & 2 \\ \end{matrix} \right],B=\left[ \begin{matrix} 2 & 3 \\ 1 & 2 \\ \end{matrix} \right]\]and\[C=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right],\] Then \[A+B-C=\left[ \begin{matrix} 2 & -1 \\ 4 & 2 \\ \end{matrix} \right]+\left[ \begin{matrix} 2 & 3 \\ 1 & 2 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 4 & 2 \\ 5 & 4 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 3 & 2 \\ 5 & 3 \\ \end{matrix} \right]\]
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