A) \[\frac{1}{2}\]
B) 2
C) 4
D) \[\frac{1}{4}\]
Correct Answer: C
Solution :
\[A=\left[ \begin{matrix} 2 & 0 \\ -a & 2 \\ \end{matrix} \right]\] \[\Rightarrow \,\,\,adjA=\,\left[ \begin{matrix} 2 & 0 \\ a & 2 \\ \end{matrix} \right]\] \[\Rightarrow \,\,{{A}^{-1}}\,=\frac{1}{4}\,\left[ \begin{matrix} 2 & 0 \\ a & 2 \\ \end{matrix} \right]\] \[\therefore \,\,\,\,\,{{A}^{-2}}={{A}^{-1}}.\,\,{{A}^{-1}}\,\,=\frac{1}{16}.\,\left[ \begin{matrix} 2 & 0 \\ a & 2 \\ \end{matrix} \right]\,\left[ \begin{matrix} 2 & 0 \\ a & 2 \\ \end{matrix} \right]\] \[=\frac{1}{16}\,.\left[ \begin{matrix} 4 & 0 \\ 4a & 4 \\ \end{matrix} \right]=\,\left[ \begin{matrix} 1/4 & 0 \\ a/4 & 1/4 \\ \end{matrix} \right]\] \[\Rightarrow \,\,x=\frac{a}{4}\] \[\therefore \,\,\,\frac{a}{x}=4\]
You need to login to perform this action.
You will be redirected in
3 sec
