A) \[\left( \begin{matrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & -\frac{1}{7} \\ \end{matrix} \right)\]
B) \[\left( \begin{matrix} -\frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & -7 \\ \end{matrix} \right)\]
C) \[\left( \begin{matrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & 7 \\ \end{matrix} \right)\]
D) \[\left( \begin{matrix} \frac{1}{3} & 0 & 0 \\ 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 7 \\ \end{matrix} \right)\]
Correct Answer: A
Solution :
Given, \[XY{{X}^{-1}}=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -7 \\ \end{matrix} \right]\] Taking inverse both the sides, we get \[{{(XY{{X}^{-1}})}^{-1}}={{\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -7 \\ \end{matrix} \right]}^{-1}}\] \[\Rightarrow \] \[X{{Y}^{-1}}{{X}^{-1}}={{\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -7 \\ \end{matrix} \right]}^{-1}}\] Now, we are to determine inverse of \[\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -7 \\ \end{matrix} \right]=A\,\,(consider)\] adj \[A=\left[ \begin{matrix} -14 & 0 & 0 \\ 0 & -21 & 0 \\ 0 & 0 & 6 \\ \end{matrix} \right]\] and \[|A|=-42\] \[{{A}^{-1}}\,=\frac{1}{|A|}\,\,(adj\,\,A)=\left[ \begin{matrix} 1/3 & 0 & 0 \\ 0 & 1/2 & 0 \\ 0 & 0 & -1/7 \\ \end{matrix} \right]\] Hence, \[X{{Y}^{-1}}{{X}^{-1}}=\left[ \begin{matrix} 1/3 & 0 & 0 \\ 0 & 1/2 & 0 \\ 0 & 0 & -1/7 \\ \end{matrix} \right]\]
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