3 and 4 .Determinants and Matrices
medium

यदि $A = \left[ {\begin{array}{*{20}{c}}0&i\\{ - i}&0\end{array}} \right]$, तो ${A^{40}}$ का मान होगा  

A

$\left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]$

B

$\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$

C

$\left[ {\begin{array}{*{20}{c}}1&1\\0&0\end{array}} \right]$

D

$\left[ {\begin{array}{*{20}{c}}{ - 1}&1\\0&{ - 1}\end{array}} \right]$

Solution

(b) $A = \left[ {\begin{array}{*{20}{c}}0&i\\{ – i}&0\end{array}} \right] \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I$

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

Standard 12
Mathematics

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