A) -1
B) 1
C) -7
D) 7
Correct Answer: C
Solution :
\[\because \] \[{{A}^{2}}\left[ \begin{matrix} 1 & 0 \\ -1 & 7 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\ -1 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ -8 & 49 \\ \end{matrix} \right]\] and \[8A+K{{l}_{2}}=8\left[ \begin{matrix} 1 & 0 \\ -1 & 7 \\ \end{matrix} \right]+K\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 8+K & 0 \\ -8 & 56+K \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{2}}=8A+K{{l}_{2}}\] \[\therefore \] \[\left[ \begin{matrix} 1 & 0 \\ -8 & 49 \\ \end{matrix} \right]=\left[ \begin{matrix} 8+K & 0 \\ -8 & 56+K \\ \end{matrix} \right]\] On comparing, we get \[8+K\text{ }=\text{ }1\] \[\therefore \] \[K=-7\]
You need to login to perform this action.
You will be redirected in
3 sec
