A) 1 only
B) 2 only
C) Both 1 only 2
D) Neither 1 nor 2
Correct Answer: D
Solution :
[d] Here, \[A=\left[ \begin{matrix} 1 & -1 \\ 2 & 3 \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix} 2 & 3 \\ -1 & -2 \\ \end{matrix} \right]\] \[\left| A \right|=3-(-2)=5\] and \[\left| B \right|=-4-(-3)=-1\] \[\Rightarrow {{A}^{-1}}=\frac{1}{5}\left[ \begin{matrix} 3 & 1 \\ -2 & 1 \\ \end{matrix} \right]\] and \[{{B}^{-1}}=-1\left[ \begin{matrix} -2 & -3 \\ 1 & 2 \\ \end{matrix} \right]\] \[AB=\left[ \begin{matrix} 3 & 5 \\ 1 & 0 \\ \end{matrix} \right]\]and \[{{A}^{-1}}{{B}^{-1}}=\frac{1}{5}\left[ \begin{matrix} 5 & 7 \\ -5 & 8 \\ \end{matrix} \right]\] \[\Rightarrow AB({{A}^{-1}}{{B}^{-1}})=\frac{1}{5}\left[ \begin{matrix} -10 & -12 \\ 5 & 7 \\ \end{matrix} \right]\ne 1\]. \[\left| AB \right|=0-5=-5\] \[\therefore {{(AB)}^{-1}}=\frac{-1}{5}\left[ \begin{matrix} 0 & -5 \\ -1 & 3 \\ \end{matrix} \right]\ne {{A}^{-1}}{{B}^{-1}}\]
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