A) \[\left[ \begin{matrix} 6 & -10 \\ 4 & 26 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} -10 & 5 \\ 4 & 24 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} -5 & -6 \\ -4 & -20 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} -5 & -7 \\ -5 & 20 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
[a] \[A=\left[ \begin{matrix} x+y & y \\ 2x & x-y \\ \end{matrix} \right]\] \[B=\left[ \begin{matrix} 2 \\ -1 \\ \end{matrix} \right]\] and \[C=\left[ \begin{matrix} 3 \\ 2 \\ \end{matrix} \right]\] Here AB = C \[\therefore \left[ \begin{matrix} x+y & y \\ 2x & x-y \\ \end{matrix} \right]\left[ \begin{matrix} 2 \\ -1 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 \\ 2 \\ \end{matrix} \right]\] \[\Rightarrow \left[ \begin{matrix} 2(x+y) & -y \\ 4x & -x+y \\ \end{matrix} \right]=\left[ \begin{matrix} 3 \\ 2 \\ \end{matrix} \right]\] \[2x+y=3...(i)\] \[3x+y=2...(ii)\] From equations (i) and (ii), we get \[x=-1\] and \[y=5\] \[\therefore A=\left[ \begin{matrix} 4 & 5 \\ -2 & -6 \\ \end{matrix} \right]\] Now, \[{{A}^{2}}=\left[ \begin{matrix} 4 & 5 \\ -2 & -6 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & 5 \\ -2 & -6 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 16-10 & 20-30 \\ -8+12 & -10+36 \\ \end{matrix} \right]=\left[ \begin{matrix} 6 & -10 \\ 4 & 26 \\ \end{matrix} \right]\]
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