3 and 4 .Determinants and Matrices
easy

यदि आव्यूह $A = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]$, तो ${A^{16}} = $

A

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

B

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

C

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

D

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

Solution

 दिया गया आव्यूह है, $A = \,\left[ {\begin{array}{*{20}{c}}0&{ – 1}\\1&0\end{array}} \right]$

हम जानते हैं,  ${A^2} = A.A = \left[ {\begin{array}{*{20}{c}}0&{ – 1}\\1&0\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}0&{ – 1}\\1&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&0\\0&{ – 1}\end{array}} \right].$               

इसलिए

${A^{16}} = {({A^2})^8} = {\left[ {\begin{array}{*{20}{c}}{ – 1}&0\\0&{ – 1}\end{array}} \right]^8} = \left[ {\begin{array}{*{20}{c}}{{{( – 1)}^8}}&0\\0&{{{( – 1)}^8}}\end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$.

Standard 12
Mathematics

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