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# 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.

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A)
Toolbox:
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 $\leq$ i $\leq$ m and 1 $\leq$ j $\leq$ n.
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• We use the principle of mathematical induction, where we need to prove P(n) is true for n=1, n=k, n=k+1
Let P(n)
To prove:
Step 1 : $A^n=\begin{bmatrix} 1+2n & -4n \\ n & 1 - 2n \end{bmatrix}$
Given $A=\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$
$\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}^n=\begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$
Substitute n=1 in the above matrix
$\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}^1=\begin{bmatrix} 1+2(1) & -4(1) \\ 1 & 1-2(1) \end{bmatrix}$
$\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}^1=\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$
Hence it is proved for n=1
Let P(n) be true for n=k
$A^k=\begin{bmatrix} 1+2k & -4k \\ k & 1 - 2k \end{bmatrix}$
Step 2:
Multiply both the sides by A
$A^k.A=\begin{bmatrix} 1+2k & -4k \\ k & 1 - 2k \end{bmatrix}A$
$A^{k+1}=\begin{bmatrix} 1+2k & -4k \\ k & 1 - 2k \end{bmatrix}\begin{bmatrix}3 &-4\\1 & -1\end{bmatrix}$
$\;\;\;=\begin{bmatrix} 3(1+2k)-4k & -4(1+2k)+4k \\ 3k+(1-2k) &-4k-( 1 - 2k) \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 3+6k-4k & -4-k+4k \\ 3k+1-2k&-4k- 1 +2k\end{bmatrix}$
$\;\;\;=\begin{bmatrix} 3+2k & -4-4k \\ k+1&-2k- 1 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 1+2(k+1)& -4(k+1) \\ k+1&1-2(k+1)\end{bmatrix}$
This prove that P(n) is true for n=k+1
Hence P(n) is true for all $n\in N$