Q.If $A=\begin{bmatrix}3 & -4\\1 & -1\end{bmatrix}$, prove that $A^n=\begin{bmatrix}1+2n & -4n\\n & 1-2n\end{bmatrix}$, for all $n\in N$.

$A=\begin{bmatrix}3 & -4\\1 & -1\end{bmatrix}$. Let $P(n): A^n=\begin{bmatrix}1+2n & -4n\\n & 1-2n\end{bmatrix}$.

Base case ($n=1$): RHS $=\begin{bmatrix}1+2 & -4\\1 & 1-2\end{bmatrix}=\begin{bmatrix}3 & -4\\1 & -1\end{bmatrix}=A$. So $P(1)$ holds.

Inductive step: Assume $P(k)$: $A^k=\begin{bmatrix}1+2k & -4k\\k & 1-2k\end{bmatrix}$. Show $P(k+1)$.

$A^{k+1}=A^k\cdot A=\begin{bmatrix}1+2k & -4k\\k & 1-2k\end{bmatrix}\begin{bmatrix}3 & -4\\1 & -1\end{bmatrix}$.

Row1: $[(1+2k)(3)+(-4k)(1),\ (1+2k)(-4)+(-4k)(-1)]=[3+6k-4k,\ -4-8k+4k]=[3+2k,\ -4-4k]$.

Row2: $[k(3)+(1-2k)(1),\ k(-4)+(1-2k)(-1)]=[3k+1-2k,\ -4k-1+2k]=[k+1,\ -2k-1]$.

So $A^{k+1}=\begin{bmatrix}2k+3 & -4k-4\\k+1 & -2k-1\end{bmatrix}$. …

Prove by the principle of mathematical induction: verify the base case $n=1$, then show the statement for $n=k$ …

• Skipping the base case verification.
• Algebra slip simplifying $A^k\cdot A$, …