A) \[{{a}^{2}}I-abA\]
B) \[{{a}^{2}}I+2abA\]
C) \[{{a}^{2}}I+{{b}^{2}}A\]
D) none of these
Correct Answer: B
Solution :
Given,\[A=\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \\ \end{matrix} \right]\] \[I=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[{{(aI+bA)}^{2}}=(aI+bA)(aI+bA)\] \[={{a}^{2}}{{I}^{2}}+aI(bA)+bA(aI)+{{(bA)}^{2}}\] \[={{a}^{2}}{{I}^{2}}2abIA+({{b}^{2}}{{A}^{2}})\] Now,\[{{I}^{2}}=I\]and\[IA=A\] \[\therefore \] \[{{(aI+bA)}^{2}}={{a}^{2}}I+2abA+{{b}^{2}}({{A}^{2}})\] Now, \[{{A}^{2}}=\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]\] \[\Rightarrow \] \[{{A}^{2}}=0\] \[\therefore \] \[{{(aI+bA)}^{2}}={{a}^{2}}I+2abA\]
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