A) -I
B) -2X
C) 2X
D) 4X
Correct Answer: C
Solution :
[c] Given matrix is: \[X=\left[ \begin{matrix} 1 & -2 \\ 0 & 3 \\ \end{matrix} \right]\] \[\therefore {{X}^{2}}=\left[ \begin{matrix} 1 & -2 \\ 0 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -2 \\ 0 & 3 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1 & -2-6 \\ 0 & 9 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -8 \\ 0 & 9 \\ \end{matrix} \right]\] So, the given expression is: \[{{X}^{4}}-2X+3I=\left[ \begin{matrix} 1 & -8 \\ 0 & 9 \\ \end{matrix} \right]-2\left[ \begin{matrix} 1 & -2 \\ 0 & 3 \\ \end{matrix} \right]+3\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1 & -8 \\ 0 & 9 \\ \end{matrix} \right]+\left[ \begin{matrix} -2 & +4 \\ 0 & -6 \\ \end{matrix} \right]+\left[ \begin{matrix} 3 & 0 \\ 0 & 3 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1-2+3 & -8+4 \\ 0 & 9-6+3 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 2 & -4 \\ 0 & 6 \\ \end{matrix} \right]=2\left[ \begin{matrix} 1 & -2 \\ 0 & 3 \\ \end{matrix} \right]=2X\]
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