A) 5
B) -1
C) 2
D) -2
Correct Answer: A
Solution :
[a] Here, \[\Rightarrow B=\frac{1}{10}\left[ \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \\ \end{matrix} \right]\] Also since, \[B={{A}^{-1}}\Rightarrow AB=I\] \[\Rightarrow \frac{1}{10}\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] \[\Rightarrow \frac{1}{10}\left[ \begin{matrix} 10 & 0 & 5-2 \\ 0 & 10 & -5+\alpha \\ 0 & 0 & 5+\alpha \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] \[\Rightarrow \frac{5-\alpha }{10}=0\Rightarrow \alpha =5\]
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