A) \[{{A}^{2}}+6I\]
B) \[I\]
C) null matrix
D) None of these
Correct Answer: C
Solution :
\[A=\left[ \begin{matrix} 4 & 2 \\ -1 & 1 \\ \end{matrix} \right],I=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[\therefore \] \[A-2I=\left[ \begin{matrix} 4 & 2 \\ -1 & 1 \\ \end{matrix} \right]-2\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 4-2 & 2-0 \\ -1-0 & 1-2 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 2 & 2 \\ -1 & -1 \\ \end{matrix} \right]\] And \[A-3I=\left[ \begin{matrix} 4 & 2 \\ -1 & 1 \\ \end{matrix} \right]-3\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 4-3 & 2 \\ -1-0 & 1-3 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1 & 2 \\ -1 & -2 \\ \end{matrix} \right]\] Now, \[(A-2I)(A-3I)=\left[ \begin{matrix} 2 & 2 \\ -1 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ -1 & -2 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 2-2 & 4-4 \\ -1+1 & -2+2 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]=\]null matrix
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