A) \[-1\]
B) \[-2\]
C) \[2\]
D) \[5\]
Correct Answer: D
Solution :
Given, \[\left[ \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \\ \end{matrix} \right]=10{{A}^{-1}}\] \[\therefore \] \[\left[ \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \\ \end{matrix} \right]\] \[\Rightarrow \] \[\left[ \begin{matrix} 10 & 0 & 0 \\ -5+\alpha & 5+\alpha & -5+\alpha \\ 0 & 0 & 10 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \\ \end{matrix} \right]\] \[\Rightarrow \] \[-5+\alpha =0\] \[\Rightarrow \] \[\alpha =5\]
You need to login to perform this action.
You will be redirected in
3 sec
