Question

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

Answer 1

It is given that $A=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

To prove: $P\left(n\right):A^n=\left[\begin{array}{cc}1+2n& -4n\\ n& 1-2n\end{array}\right],n\in N$

We shall prove the result by using the principle of mathematical induction .

For $n=1$, we have:

$P\left(1\right):A^1=\left[\begin{array}{cc}1+2& -4\\ 1& 1-2\end{array}\right]=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]=A$

Therefore, the result is true for $n=1$.

Let the result be true for $n=k$.

That is ,
$P\left(k\right):A^k=\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right],n\in N$

Now, we prove that the result is true for $n=k+1$

Consider

$A^{k+1}=A^k.A$

$=\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

$=\left[\begin{array}{cc}3(1+2k)-4k& -4(1+2k)+4k\\ 3k+1-2k& -4k-1(1-2k)\end{array}\right]$

$=\left[\begin{array}{cc}3+6k-4k& -4-4k\\ 3k+1-2k& -4k-1+2k\end{array}\right]$

$=\left[\begin{array}{cc}3+2k& -4-4k\\ 1+k& -1-2k\end{array}\right]$

$=\left[\begin{array}{cc}1+2(k+1)& -4(k+1)\\ 1+k& 1-2(k+1)\end{array}\right]$

Therefore, the result is true for $n=k+1$.
Thus, by the principle of mathematical induction , we have:

$A^n=\left[\begin{array}{cc}1+2n& -4n\\ n& 1-2n\end{array}\right],n\in N$