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

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