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यदि $A = \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]$, तो ${A^n} = $
$\left[ {\begin{array}{*{20}{c}}1&n\\0&1\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}n&n\\0&n\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}n&1\\0&n\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}1&1\\0&n\end{array}} \right]$
Solution
(a) ${A^2} = \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right],$
और ${A^3} = {A^2}.A = \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right]$
==> ${A^n} = {A^{n – 1}}.A = \left[ {\begin{array}{*{20}{c}}1&{n – 1}\\0&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&n\\0&1\end{array}} \right]$.