If A = [ * 20 c 1&1 1&1 ],then A^ 100 = - Vidyadip

MCQ

If $A = \left[ {\begin{array}{*{20}{c}}1&1\\1&1\end{array}} \right],$then ${A^{100}} = $

A
${2^{100}}A$
✓
${2^{99}}A$
C
${2^{101}}A$
D
None of these

✓

Answer

Correct option: B.

${2^{99}}A$

b
(b) $A = \left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right]$

${A^2} = \left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right]\,\left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right]$= $\left[ {\,\begin{array}{*{20}{c}}2&2\\2&2\end{array}\,} \right] = 2\left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right]$

${A^3} = 2\,\left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right]\,\left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right] = {2^2}\left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right]$

${A^n} = {2^{n - 1}}\left[ {\,\begin{array}{*{20}{c}}1&1\\1&1\end{array}\,} \right]$ ==> ${A^{100}} = {2^{99}}\left[ {\begin{array}{*{20}{c}}1&1\\1&1\end{array}} \right]$.

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