- Home
- Standard 12
- Mathematics
यदि $A = \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right],$तो ${A^n} = $
$\left[ {\begin{array}{*{20}{c}}1&{2n}\\0&1\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}2&n\\0&1\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}1&{2n}\\0&{ - 1}\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}1&{2n}\\1&0\end{array}} \right]$
Solution
(a) ${A^2} = \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&4\\0&1\end{array}} \right]$ तथा ${A^3} = {A^2}A$.
$\Rightarrow$ $\left[ {\,\begin{array}{*{20}{c}}1&4\\0&1\end{array}\,} \right]\,\,\left[ {\,\begin{array}{*{20}{c}}1&2\\0&1\end{array}\,} \right] = \left[ {\,\begin{array}{*{20}{c}}1&6\\0&1\end{array}\,} \right]$ and so on.
$\therefore $ ${A^n} = \left[ {\,\begin{array}{*{20}{c}}1&{2n}\\0&1\end{array}\,} \right]$.