Q.Prove by method of induction $\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}^n = \begin{pmatrix} 2n+1 & -4n \\ n & -2n+1 \end{pmatrix}$, $\forall n \in N$.

Let $A=\begin{pmatrix}3&-4\1&-1\end{pmatrix}$, $P(n):A^n=\begin{pmatrix}2n+1&-4n\n&-2n+1\end{pmatrix}$. Base: $n=1$: R.H.S.$=\begin{pmatrix}3&-4\1&-1\end{pmatrix}=A$; holds. Hypothesis: assume $A^k=\begin{pmatrix}2k+1&-4k\k&-2k+1\end{pmatrix}$. Step: $A^{k+1}=A^k A=\begin{pmatrix}2k+1&-4k\k&-2k+1\end{pmatrix}\begin{pmatrix}3&-4\1&-1\end{pmatrix}$. Entry(1,1)$=3(2k+1)-4k=6k+3-4k=2k+3=2(k+1)+1$. Entry(1,2)$=-4(2k+1)+4k=-8k-4+4k=-4k-4=-4(k+1)$. Entry(2,1)$=3k+(-2k+1)=k+1$. Entry(2,2)$=-4k-(-2k+1)=-4k+2k-1=-2k-1=-2(k+1)+1$. So $A^{k+1}=\begi …

Multiply assumed matrix by $A$ using standard 2x2 matrix multiplication, s …