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