Question
If \[A = \left[ {\begin{array}{*{20}{c}}
2&2\\
2&2
\end{array}} \right],\] then what is $$A^n$$ equal to ?
A.
\[\left[ {\begin{array}{*{20}{c}}
{{2^n}}&{{2^n}}\\
{{2^n}}&{{2^n}}
\end{array}} \right]\]
B.
\[\left[ {\begin{array}{*{20}{c}}
2n&2n\\
2n&2n
\end{array}} \right]\]
C.
\[\left[ {\begin{array}{*{20}{c}}
{{2^{2n - 1}}}&{{2^{2n - 1}}}\\
{{2^{2n - 1}}}&{{2^{2n - 1}}}
\end{array}} \right]\]
D.
\[\left[ {\begin{array}{*{20}{c}}
{{2^{2n + 1}}}&{{2^{2n + 1}}}\\
{{2^{2n + 1}}}&{{2^{2n + 1}}}
\end{array}} \right]\]
Answer :
\[\left[ {\begin{array}{*{20}{c}}
{{2^{2n - 1}}}&{{2^{2n - 1}}}\\
{{2^{2n - 1}}}&{{2^{2n - 1}}}
\end{array}} \right]\]
Solution :
Given matrix is :
\[\begin{array}{l}
A = \left[ {\begin{array}{*{20}{c}}
2&2\\
2&2
\end{array}} \right]\\
{A^2} = \left[ {\begin{array}{*{20}{c}}
2&2\\
2&2
\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}
2&2\\
2&2
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{4 + 4}&{4 + 4}\\
{4 + 4}&{4 + 4}
\end{array}} \right]\\
= \,\left[ {\begin{array}{*{20}{c}}
{{2^3}}&{{2^3}}\\
{{2^3}}&{{2^3}}
\end{array}} \right]\\
{A^3} = \left[ {\begin{array}{*{20}{c}}
8&8\\
8&8
\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}
2&2\\
2&2
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{16 + 16}&{16 + 16}\\
{16 + 16}&{16 + 16}
\end{array}} \right]\\
= \,\left[ {\begin{array}{*{20}{c}}
{32}&{32}\\
{32}&{32}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{2^5}}&{{2^5}}\\
{{2^5}}&{{2^5}}
\end{array}} \right]
\end{array}\]
Going this way, we get,
\[\begin{array}{l}
{A^4} = \left[ {\begin{array}{*{20}{c}}
{{2^7}}&{{2^7}}\\
{{2^7}}&{{2^7}}
\end{array}} \right]\\
\Rightarrow \,{A^n} = \left[ {\begin{array}{*{20}{c}}
{{2^{2n - 1}}}&{{2^{2n - 1}}}\\
{{2^{2n - 1}}}&{{2^{2n - 1}}}
\end{array}} \right]
\end{array}\]