3 and 4 .Determinants and Matrices
easy

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

A

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

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

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

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

Standard 12
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.