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