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