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જો $A = \left[ {\begin{array}{*{20}{c}}1&1\\1&1\end{array}} \right],$ તો ${A^{100}} = $
${2^{100}}A$
${2^{99}}A$
${2^{101}}A$
એકપણ નહી.
Solution
(b) $A = \left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right]$
${A^2} = \left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right]\,\left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right]$= $\left[ {\,\begin{array}{*{20}{c}}2&2\\2&2\end{array}\,} \right] = 2\left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right]$
${A^3} = 2\,\left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right]\,\left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right] = {2^2}\left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right]$
${A^n} = {2^{n – 1}}\left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right]$ ==> ${A^{100}} = {2^{99}}\left[ {\begin{array}{*{20}{c}}1&1\\1&1\end{array}} \right]$.
