Question

If A = $\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$ then $A^k=\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right]$
where k is any +ve integer .

Answer 1

A = $\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$ then $A^1={\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right]}_{k=1}$
Also $A^2$ = AA
=$\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$ $\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$ = $\left[\begin{array}{cc}9-4& -12+4\\ 3-1& -4+1\end{array}\right]$
or $A^2$ = $\left[\begin{array}{cc}5& -8\\ 2& -3\end{array}\right]$ = ${\left[\begin{array}{cc}1+22& -4.2\\ 2& 1-2.2\end{array}\right]}_{k=2}$
Now assume that $A^k={\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right]}_{k=k}$
$\therefore A^{k+1}=A{A}^k=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right]$
= $\left[\begin{array}{cc}3+6k-4& -12k-4+8k\\ 1+2k-k& -4k-1+2k\end{array}\right]$
= $\left[\begin{array}{cc}3+2k& -4k-4\\ 1+k& -2k-1\end{array}\right]$
= $\left[\begin{array}{cc}1+2(k+1)& -4(k+1)\\ 1+k& 1-2k+1\end{array}\right]$
we observe that our assumption is true for k = k + 1 and it was true when k = 1 or 2. Hence it is true for all +ve integral values for k .