A) 1
B) 0
C) 2
D) None of these
Correct Answer: B
Solution :
[b] We have, \[I=A{{A}^{-1}}\] \[=\frac{1}{2}\left[ \begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -1 & 1 \\ -8 & 6 & 2c \\ 5 & -3 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1 & 0 & c+1 \\ 0 & 1 & 2(c+1) \\ 4(1-a) & 3(a-1) & 2+ac \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] Comparing the elements we get \[c+1=0\Rightarrow c=-1\] and \[a-1=0\Rightarrow a=1\]
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