Question

lf $n$ is a natural number $A=\left\{\begin{array}{cc}5& -8\\ 2& -3\end{array}\right\}$ then $A^n=$

• A. $\left(\begin{array}{cccc}6-& n& -2n& -6\\ 2n& & 1-4n& \end{array}\right)$
• B. $\left(\begin{array}{cc}4+n& -8n\\ 2n& 2-5n\end{array}\right)$
• C. $\left(\begin{array}{cc}1+4n& -8n\\ 2n& 1-4n\end{array}\right)$
• D. $\left(\begin{array}{ccc}6-& n& -8n\\ 2n& & 1-4n\end{array}\right)$

Answer 1

The correct option is C $\left(\begin{array}{cc}1+4n& -8n\\ 2n& 1-4n\end{array}\right)$

$A=\left[\begin{array}{cc}5& -8\\ 2& -3\end{array}\right]$
$A=\left[\begin{array}{cc}1+4& -8\\ 2& 1-4\end{array}\right]$
$=I+\left[\begin{array}{cc}4& -8\\ 2& -4\end{array}\right]\downarrow B$
$A^n=(I+B{)}^n$
$=I^n{+}_{n-1}^n\text{7^{n-1}B}{+}_{n-2}^n\text{I^{n-1}B^2}+B^n$
$B^2=\left[\begin{array}{cc}4& -8\\ 2& -4\end{array}\right]\left[\begin{array}{cc}4& -8\\ 2& -4\end{array}\right]$
$=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$\therefore A^n=I^n+n^B=I+nB$
$=\left[\begin{array}{cc}1+4n& -8n\\ 2n& 1-4n\end{array}\right]$