Let $P(n): A^n=\left[\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right]$
where $\mathrm{A}=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$
Put $n=1 \quad A=\left[\begin{array}{cc}1+2 & -4 \\ 1 & 1-2\end{array}\right]=\left[\begin{array}{cc}3 & -4 \\ 1 & -1\end{array}\right]$
$\therefore \quad \mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=1$
Let $P(n)$ be true for $n=K$
$$
\therefore \mathrm{A}^{\mathrm{K}}=\left[\begin{array}{cc}
1+2 \mathrm{~K} & -4 \mathrm{~K} \\
\mathrm{~K} & 1-2 \mathrm{~K}
\end{array}\right]
$$
Multiplying both sides by $\mathrm{A}$
$\mathrm{A}^{\mathrm{K}} \mathrm{A}=\mathrm{A}^{\mathrm{K}+1}=\left[\begin{array}{cc}1+2 \mathrm{~K} & -4 \mathrm{~K} \\ \mathrm{~K} & 1-2 \mathrm{~K}\end{array}\right] \mathrm{A}$
$=\left[\begin{array}{cc}1+2 \mathrm{~K} & -4 \mathrm{~K} \\ \mathrm{~K} & 1-2 \mathrm{~K}\end{array}\right]\left[\begin{array}{cc}3 & -4 \\ 1 & -1\end{array}\right]$
$=\left[\begin{array}{cc}3+2 \mathrm{~K} & -4-4 \mathrm{~K} \\ 1+\mathrm{K} & -1-2 \mathrm{~K}\end{array}\right]=\left[\begin{array}{cc}1+2(\mathrm{~K}+1) & -4(\mathrm{~K}+1) \\ \mathrm{K}+1 & 1-2(\mathrm{~K}+1)\end{array}\right]$
This proves that $\mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=\mathrm{K}+1$. Hence, $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{N}$.