Question

If $A=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right]$ then $A^n=\dots \dots \dots .\dots ,n\in N$

• A. $I$
• B. $A$
• C. $1/2A$
• D. $2A$

Answer 1

Answer: (B)

$A$

$A=\frac{1}{2}\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$

$A^2=\frac{1}{4}\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$

$=\frac{1}{4}\left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$

$A^2=A$

same $A^3=\frac{1}{4}\left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]=A$

$\therefore A^n=A$