A) \[a=1=2b\]
B) \[a=b\]
C) \[a={{b}^{2}}\]
D) \[ab=1\]
Correct Answer: D
Solution :
\[{{\left[ \begin{matrix} 0 & a \\ b & a \\ \end{matrix} \right]}^{4}}=I\] Now, \[{{\left[ \begin{matrix} 0 & a \\ b & 0 \\ \end{matrix} \right]}^{2}}=\left[ \begin{matrix} 0 & a \\ b & 0 \\ \end{matrix} \right]\,\,\left[ \begin{matrix} 0 & a \\ b & 0 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} ab & 0 \\ 0 & ab \\ \end{matrix} \right]\] and \[={{\left[ \begin{matrix} 0 & a \\ b & 0 \\ \end{matrix} \right]}^{4}}={{\left[ \begin{matrix} 0 & a \\ b & 0 \\ \end{matrix} \right]}^{2}}{{\left[ \begin{matrix} 0 & a \\ b & 0 \\ \end{matrix} \right]}^{2}}\] \[=\left[ \begin{matrix} ab & a \\ 0 & ab \\ \end{matrix} \right]\,\,\left[ \begin{matrix} ab & 0 \\ 0 & ab \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} {{c}^{2}}{{b}^{2}} & a \\ 0 & {{a}^{2}}{{b}^{2}} \\ \end{matrix} \right]\] But \[{{\left[ \begin{matrix} 0 & a \\ b & 0 \\ \end{matrix} \right]}^{4}}\,\,=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] (given) \[\Rightarrow \]\[\left[ \begin{matrix} {{a}^{2}}{{b}^{2}} & 0 \\ 0 & {{a}^{2}}{{b}^{2}} \\ \end{matrix} \right]\,\,=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[\Rightarrow \] \[{{a}^{2}}{{b}^{2}}=1\] \[\Rightarrow \] \[ab=1\]
You need to login to perform this action.
You will be redirected in
3 sec
