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

MCQ

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

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

✓

Answer

Correct option: A.

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

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

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

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

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