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Mathematics Question on Matrices

If A=[3−4\1−1]A=\begin{bmatrix}3&-4\\\1&-1\end{bmatrix},then prove An=[1+2n−4n n1−2n]A_n=\begin{bmatrix}1+2n& -4n\\\ n& 1-2n\end{bmatrix}where n is any positive integer

Answer

It is given that A=[3−4\1−1]A=\begin{bmatrix}3&-4\\\1&-1\end{bmatrix}
To prove:P(n):An=[1+2n−4n n1−2n]n∈NP(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)=[1+2(1)−4(1) 11−2(1)]P(1)=\begin{bmatrix}1+2(1)& -4(1)\\\ 1& 1-2(1)\end{bmatrix}
A=[3−4\1−1]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):Ak=[1+2k−4k k1−2k]n∈NP(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
Ak+1=Ak.AA^{k+1}=A^k.A
[1+2k−4k k1−2k][3−4\1−1]\begin{bmatrix}1+2k& -4k\\\ k& 1-2k\end{bmatrix}\begin{bmatrix}3&-4\\\1&-1\end{bmatrix}
=[3+6k−4k−4−8k+4k 3k+1−2k−4k−1+2k]=\begin{bmatrix}3+6k-4k& -4-8k+4k\\\ 3k+1-2k& -4k-1+2k\end{bmatrix}
=[3+2k−4−4k 1+k−1−2k]=\begin{bmatrix}3+2k& -4-4k\\\ 1+k& -1-2k\end{bmatrix}
=[1+2(k+1)−4(k+1) (k+1)1−2(k+1)]= \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+1n = k + 1.
Thus, by the principle of mathematical induction, we have:An=[1+2n−4n n1−2n]n∈NA^n=\begin{bmatrix}1+2n& -4n\\\ n& 1-2n\end{bmatrix}n\in N