If A = [ * 20 c 1&2 0&1 ],then A^n = - Vidyadip

MCQ

If $A = \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right],$then ${A^n} = $

✓
$\left[ {\begin{array}{*{20}{c}}1&{2n}\\0&1\end{array}} \right]$
B
$\left[ {\begin{array}{*{20}{c}}2&n\\0&1\end{array}} \right]$
C
$\left[ {\begin{array}{*{20}{c}}1&{2n}\\0&{ - 1}\end{array}} \right]$
D
$\left[ {\begin{array}{*{20}{c}}1&{2n}\\1&0\end{array}} \right]$

✓

Answer

Correct option: A.

$\left[ {\begin{array}{*{20}{c}}1&{2n}\\0&1\end{array}} \right]$

a
(a) ${A^2} = \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&4\\0&1\end{array}} \right]$ and ${A^3} = {A^2}A$.

= $\left[ {\,\begin{array}{*{20}{c}}1&4\\0&1\end{array}\,} \right]\,\,\left[ {\,\begin{array}{*{20}{c}}1&2\\0&1\end{array}\,} \right] = \left[ {\,\begin{array}{*{20}{c}}1&6\\0&1\end{array}\,} \right]$ and so on.

$\therefore $ ${A^n} = \left[ {\,\begin{array}{*{20}{c}}1&{2n}\\0&1\end{array}\,} \right]$.

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