Question

If A = $\left[\begin{array}{c}1\\ 1/2\end{array}\begin{array}{c}0\\ 1\end{array}\right]$, then ${\text{A}}^{100}$ is equal to

• A. $\left[\begin{array}{c}1\\ 25\end{array}\begin{array}{c}0\\ 0\end{array}\right]$
• B. $\left[\begin{array}{c}1\\ 50\end{array}\begin{array}{c}0\\ 1\end{array}\right]$
• C. $\left[\begin{array}{c}1\\ {\left(1/2\right)}^{100}\end{array}\begin{array}{c}0\\ 1\end{array}\right]$
• D. None of these

Answer 1

The correct option is B $\left[\begin{array}{c}1\\ 50\end{array}\begin{array}{c}0\\ 1\end{array}\right]$

$A=\left[\begin{array}{cc}1& 0\\ \frac{1}{2}& 1\end{array}\right]$

$A^2=AA=\left[\begin{array}{cc}1& 0\\ \frac{1}{2}& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ \frac{1}{2}& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ \frac{1}{2}+\frac{1}{2}& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$

$A^4=A^2{A}^2=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 1+1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]$

$A^8=A^4{A}^4=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 2+2& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 4& 1\end{array}\right]$

$A^{1}6=A^8{A}^8=\left[\begin{array}{cc}1& 0\\ 4& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 4& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 4+4& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 8& 1\end{array}\right]$

we notice a pattern i.e $A^{2^n}=\left[\begin{array}{cc}1& 0\\ 2^{n-1}& 1\end{array}\right]$

$\therefore A^{100}=A^{64}{A}^{32}{A}^4$

$=\left[\begin{array}{cc}1& 0\\ 32& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 16& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]$

$=\left[\begin{array}{cc}1& 0\\ 32+16& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]$

$=\left[\begin{array}{cc}1& 0\\ 48& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]$

$=\left[\begin{array}{cc}1& 0\\ 48+2& 1\end{array}\right]$

$=\left[\begin{array}{cc}1& 0\\ 50& 1\end{array}\right]$