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3 and 4 .Determinants and Matrices
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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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