Question

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

Answer 1

We are required to prove that for all $n\in N.$

$P\left(n\right)=A^n=\left[\begin{array}{cc}1+2n& -4n\\ n& 1-2n\end{array}\right]$
Let n =1
$P\left(1\right):A^1=\left[\begin{array}{cc}1+2\left(1\right)& -4\left(1\right)\\ 1& 1-2\left(1\right)\end{array}\right]=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right].........\left(i\right)$
$\therefore$ which is true for n=1.
Let the result be true for n=k.
$P\left(k\right)=A^k=\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right].....\left(ii\right)$
Let n =k+1
Then, $P(k+1):A^{k+1}=\left[\begin{array}{cc}1+2(k+1)& -4(k+1)\\ k+1& 1-2(k+1)\end{array}\right]=\left[\begin{array}{cc}2k+3& -4k+4\\ k+1& -2k-1\end{array}\right]$
Now, $LHS=A^{k+1}=A^k{A}^I=\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$ [Using Eqs. (i)and (ii)]
$=\left[\begin{array}{cc}(1+2k).3+(-4k).1& (1+2k).(-4)+(-4k).(-1)\\ k.3+(1-2k).1& k.(-4)+(1-2k).(-1)\end{array}\right]=\left[\begin{array}{cc}3+2k& -4-4k\\ k+1& -1-2k\end{array}\right]=\left[\begin{array}{cc}1+2(k+1)& -4(k+1)\\ k+1& 1-2(k+1)\end{array}\right]$
Therefore, the result is true for n=k+1 whenever it is true for n=k. So, by principle of mathematical induction, it is true for all $n\in N.$