A) \[\left[ \begin{matrix} 1 & 7 \\ 1 & 7 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 2 & 6 \\ 0 & 8 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 2 & 6 \\ 0 & 6 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 2 & 6 \\ 0 & 7 \\ \end{matrix} \right]\]
Correct Answer: B
Solution :
[b] Given that, \[A=\left[ \begin{matrix} 1 & 2 \\ 0 & 3 \\ \end{matrix} \right]\] \[{{A}^{2}}=\left[ \begin{matrix} 1 & 2 \\ 0 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ 0 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 2+6 \\ 0 & 9 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 8 \\ 0 & 9 \\ \end{matrix} \right]\] Since, \[f(x)={{x}^{2}}-x+2\] Putting A in place of x \[f(A)={{A}^{2}}-A+2I\] \[=\left[ \begin{matrix} 1 & 8 \\ 0 & 9 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 2 \\ 0 & 3 \\ \end{matrix} \right]+\left[ \begin{matrix} 2 & 0 \\ 0 & 2 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1-1+2 & 8-2+0 \\ 0-0+0 & 9-3+2 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 2 & 6 \\ 0 & 8 \\ \end{matrix} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
