A) \[2014\]
B) \[2016\]
C) \[-175\]
D) \[- 25\]
Correct Answer: B
Solution :
) \[A=\left[ \begin{matrix} -4 & -1 \\ 3 & 1 \\ \end{matrix} \right]\] \[{{A}^{2}}=\left[ \begin{matrix} -4 & -1 \\ 3 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} -4 & -1 \\ 3 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 13 & 3 \\ -9 & -2 \\ \end{matrix} \right]\] \[{{A}^{2}}-2A-I=\left[ \begin{matrix} 13 & 3 \\ -9 & -2 \\ \end{matrix} \right]-\left[ \begin{matrix} -8 & -2 \\ 6 & 2 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 20 & 5 \\ -15 & 5 \\ \end{matrix} \right]\]And \[\left| A \right|=-1\] \[\Rightarrow \,\,\left| {{A}^{2016}}-2{{A}^{2014}} \right|={{\left| A \right|}^{2014}}\] \[\left| {{A}^{2}}-2A-I \right|=1\left| \begin{matrix} 20 & 5 \\ -15 & -5 \\ \end{matrix} \right|=(-100+75)=-25\]
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