A) \[k=7\]
B) \[k=-7\]
C) \[k=0\]
D) None of these
Correct Answer: B
Solution :
[b] We have, \[{{A}^{2}}=\left[ \begin{matrix} 1 & 0 \\ -1 & 7 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\ -1 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ -8 & 49 \\ \end{matrix} \right]\] and \[8A+kI=8\left[ \begin{matrix} 1 & 0 \\ -1 & 7 \\ \end{matrix} \right]+k\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 8 & 0 \\ -8 & 56 \\ \end{matrix} \right]+\left[ \begin{matrix} k & 0 \\ 0 & k \\ \end{matrix} \right]=\left[ \begin{matrix} 8+k & 0 \\ -8 & 56+k \\ \end{matrix} \right]\] Thus, \[{{A}^{2}}=8A+kI\Rightarrow \left[ \begin{matrix} 1 & 0 \\ -8 & 49 \\ \end{matrix} \right]=\left[ \begin{matrix} 8+k & 0 \\ -8 & 56+k \\ \end{matrix} \right]\] \[\Rightarrow 1=8+k\] and \[56+k=49\Rightarrow k=-7\]
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