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# If $\begin{bmatrix}0&a\\b&0\end{bmatrix}^4=I$ then

$(a)\;a=1=2b\qquad(b)\;a=b\qquad(c)\;a=b^2\qquad(d)\;ab=1$

Can you answer this question?

Given :
$\begin{bmatrix}0&a\\b&0\end{bmatrix}^4=I$
$\begin{bmatrix}0 &a\\b&0\end{bmatrix}^2=\begin{bmatrix}0 &a\\b&0\end{bmatrix}\begin{bmatrix}0 &a\\b&0\end{bmatrix}$
$\qquad\qquad=\begin{bmatrix}ab&0\\0&ab\end{bmatrix}$
$\begin{bmatrix}0 &a\\b&0\end{bmatrix}^4=\begin{bmatrix}0 &a\\b&0\end{bmatrix}^2\begin{bmatrix}0 &a\\b&0\end{bmatrix}^2$
$\qquad\qquad=\begin{bmatrix}ab&a\\0&ab\end{bmatrix}\begin{bmatrix}ab&a\\0&ab\end{bmatrix}$
$\qquad\qquad=\begin{bmatrix}a^2b^2&0\\0&a^2b^2\end{bmatrix}$
$\begin{bmatrix}0 &a\\b&0\end{bmatrix}^4=\begin{bmatrix}1 &0\\0&1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}a^2b^2&0\\0&a^2b^2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow a^2b^2=1$
$\Rightarrow ab=1$
Hence (d) is the correct answer.
answered Nov 25, 2013