MCQ
If $A = \left[ {\begin{array}{*{20}{c}}0&i\\{ - i}&0\end{array}} \right]$, then the value of ${A^{40}}$ is
  • A
    $\left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]$
  • $\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$
  • C
    $\left[ {\begin{array}{*{20}{c}}1&1\\0&0\end{array}} \right]$
  • D
    $\left[ {\begin{array}{*{20}{c}}{ - 1}&1\\0&{ - 1}\end{array}} \right]$

Answer

Correct option: B.
$\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$
b
(b) $A = \left[ {\begin{array}{*{20}{c}}0&i\\{ - i}&0\end{array}} \right] \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I$

==> ${({A^2})^{20}} = {A^{40}} = {(I)^{20}} = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$.

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