Question

If \(A=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}\),then prove \(A_n=\begin{bmatrix}1+2n& -4n\\ n& 1-2n\end{bmatrix}\)where n is any positive integer

Answer 1

It is given that \(A=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}\)
To prove:\(P(n):A^n=\begin{bmatrix}1+2n& -4n\\ n& 1-2n\end{bmatrix}n\in N\)
We shall prove the result by using the principle of mathematical induction.
For n = 1, we have:
\(P(1)=\begin{bmatrix}1+2(1)& -4(1)\\ 1& 1-2(1)\end{bmatrix}\)
\(A=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}\)
Therefore, the result is true for n = 1.
Let the result be true for n = k.
That is,
\(P(k):A^k=\begin{bmatrix}1+2k& -4k\\ k& 1-2k\end{bmatrix}n\in N\)
Now, we prove that the result is true for n = k + 1.
Consider 
\(A^{k+1}=A^k.A\)
\(\begin{bmatrix}1+2k& -4k\\ k& 1-2k\end{bmatrix}\begin{bmatrix}3&-4\\1&-1\end{bmatrix}\)
\(=\begin{bmatrix}3+6k-4k& -4-8k+4k\\ 3k+1-2k& -4k-1+2k\end{bmatrix}\)
\(=\begin{bmatrix}3+2k& -4-4k\\ 1+k& -1-2k\end{bmatrix}\)
\(= \begin{bmatrix}1+2(k+1)& -4(k+1)\\ (k+1)& 1-2(k+1)\end{bmatrix}\)
Therefore, the result is true for \(n = k + 1\).
Thus, by the principle of mathematical induction, we have:\(A^n=\begin{bmatrix}1+2n& -4n\\ n& 1-2n\end{bmatrix}n\in N\)