A) \[\left[ \begin{matrix} 4 & -32 \\ 0 & 36 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 4 & 0 \\ -32 & 36 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 36 & 0 \\ -32 & 4 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 36 & -32 \\ 0 & 4 \\ \end{matrix} \right]\]
Correct Answer: D
Solution :
A is a matrix such that \[A\cdot \left[ \begin{matrix} 1 & 2 \\ 0 & 3 \\ \end{matrix} \right]\] is a scalar matrix and \[|3A|=108\] Let the scalar matrix be \[\left[ \begin{matrix} k & 0 \\ 0 & k \\ \end{matrix} \right]\] \[\Rightarrow A\cdot \left[ \begin{matrix} 1 & 2 \\ 0 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} k & 0 \\ 0 & k \\ \end{matrix} \right]\] \[\Rightarrow A=\left[ \begin{matrix} k & 0 \\ 0 & k \\ \end{matrix} \right]{{\left[ \begin{matrix} 1 & 2 \\ 0 & 3 \\ \end{matrix} \right]}^{-1}}.....[\because AB=C\Rightarrow A=C{{B}^{-1}}]\] Let \[B=\left[ \begin{matrix} 1 & 2 \\ 0 & 3 \\ \end{matrix} \right]\] Now, \[|B|=3\] Then, \[{{B}^{-1}}=\frac{1}{|B|}Co-factor\] matrix of B \[\Rightarrow A=\frac{1}{3}\left[ \begin{matrix} k & 0 \\ 0 & k \\ \end{matrix} \right]\left[ \begin{matrix} 3 & -2 \\ 0 & 1 \\ \end{matrix} \right]\] \[\Rightarrow A=\left[ \begin{matrix} k & 0 \\ 0 & k \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -\frac{2}{3} \\ 0 & \frac{1}{3} \\ \end{matrix} \right]\] \[\Rightarrow A=\left[ \begin{matrix} k & -\frac{2}{3}k \\ 0 & \frac{k}{3} \\ \end{matrix} \right].........(i)\] \[|3A|=108...........[Given]\] \[\Rightarrow 108=|3A|=3|A|=\left| \begin{matrix} 3k & -2k \\ 0 & k \\ \end{matrix} \right|\] \[\Rightarrow 3{{k}^{2}}=108\] \[\Rightarrow {{k}^{2}}=36\] \[\Rightarrow k=\pm 6\] Take \[k=6\] \[\Rightarrow A=\left[ \begin{matrix} 6 & -4 \\ 0 & 2 \\ \end{matrix} \right]..........From(i)\] \[\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 6 & -4 \\ 0 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 6 & -4 \\ 0 & 2 \\ \end{matrix} \right]\] \[\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 36 & -32 \\ 0 & 4 \\ \end{matrix} \right]\] For \[k=-6\] \[\Rightarrow A=\left[ \begin{matrix} -6 & 4 \\ 0 & -2 \\ \end{matrix} \right]......From\,\,(i)\] \[\Rightarrow {{A}^{2}}=\left[ \begin{matrix} -6 & 4 \\ 0 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} -6 & 4 \\ 0 & -2 \\ \end{matrix} \right]\] \[\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 36 & -32 \\ 0 & 4 \\ \end{matrix} \right]\] Hence, \[{{A}^{2}}=\left[ \begin{matrix} 36 & -32 \\ 0 & 4 \\ \end{matrix} \right]\]
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