A) 1 or -1
B) 5 or -1
C) 5 or 1
D) no real values
Correct Answer: B
Solution :
[b] We have \[{{A}^{2}}=\left[ \begin{matrix} \alpha & 0 \\ 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} \alpha & 0 \\ 1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} {{\alpha }^{2}} & 0 \\ \alpha +1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 9 & a \\ b & c \\ \end{matrix} \right]\] \[\Rightarrow \] we get \[{{\alpha }^{2}}=9\Rightarrow \alpha =\pm 3\] and \[a=0,c=1,b=\alpha +1=3+1=4\] or \[b=-3+1=-2\] So \[a+b+c=(0+4+1)=5\] or \[(0-2+1)=-1\]
You need to login to perform this action.
You will be redirected in
3 sec
