Question

If $A=\left[\begin{array}{cc}1& 0\\ 1& 2\end{array}\right]$, then $A^{32}$ is equal to

• A. $\left[\begin{array}{cc}1& 0\\ 2^{32}+1& 2^{32}\end{array}\right]$
• B. $\left[\begin{array}{cc}1& 0\\ 2^{32}& 2^{32}\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 0\\ 2^{32}-1& 2^{32}\end{array}\right]$
• D. $\left[\begin{array}{cc}{2}^{32}& 0\\ 2^{32}& 2^{33}\end{array}\right]$

Answer 1

The correct option is C $\left[\begin{array}{cc}1& 0\\ 2^{32}-1& 2^{32}\end{array}\right]$

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

$A=I+B$
where $I$ is the identity matrix and $B=\left[\begin{array}{cc}0& 0\\ 1& 1\end{array}\right]$

$B^2=B$
$\Rightarrow B$ is an idempotent matrix.
So, $B=B^2=B^3=\cdots =B^{32}$

Now, $A^{32}=(I+B{)}^{32}$
$\Rightarrow A^{32}=I+{}^{32}{C}_{1}B+{}^{32}{C}_2{B}^2+{}^{32}{C}_3{B}^3+\cdots +{}^{32}{C}_{32}{B}^{32}$
$\Rightarrow A^{32}=I+{}^{32}{C}_{1}B+{}^{32}{C}_{2}B+{}^{32}{C}_{3}B+\cdots +{}^{32}{C}_{32}B$
$\Rightarrow A^{32}=I+({}^{32}{C}_1+{}^{32}{C}_2+\cdots +{}^{32}{C}_{32})B$
$\Rightarrow A^{32}=I+(2^{32}-1)B$

$\therefore A^{32}=\left[\begin{array}{cc}1& 0\\ 2^{32}-1& 2^{32}\end{array}\right]$