A) \[-2\]
B) \[-1\]
C) 2
D) 5
Correct Answer: D
Solution :
Since, inverse of matrix A is B, ie,\[B={{A}^{-1}}\] \[\therefore \] \[10{{A}^{-1}}=\left[ \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \\ \end{matrix} \right]\] \[10{{A}^{-1}}A=\left[ \begin{matrix} 4 & 2 & 3 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \\ \end{matrix} \right]A\] \[\Rightarrow \] \[10I=\left[ \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \\ \end{matrix} \right]\] \[\Rightarrow \]\[\left[ \begin{matrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \\ \end{matrix} \right]=\left[ \begin{matrix} 10 & 0 & 0 \\ -5+\alpha & 5+\alpha & -5+\alpha \\ 0 & 0 & 10 \\ \end{matrix} \right]\] \[\Rightarrow \] \[-5+\alpha =0\] \[\Rightarrow \] \[\alpha =5\]
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