3 and 4 .Determinants and Matrices
easy

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

A

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

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

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

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

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

Std 12
Mathematics

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