# If, then prove where n is any positive integer

Question:

If $A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$, then prove $A^{n}=\left[\begin{array}{ll}1+2 n & -4 n \\ n & 1-2 n\end{array}\right]$ where $n$ is any positive integer

Solution:

It is given that $A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$

To prove: $\quad \mathrm{P}(n): A^{n}=\left[\begin{array}{ll}1+2 n & -4 n \\ n & 1-2 n\end{array}\right], n \in \mathbf{N}$

We shall prove the result by using the principle of mathematical induction.

For n = 1, we have:

$P(1): A^{1}=\left[\begin{array}{ll}1+2 & -4 \\ 1 & 1-2\end{array}\right]=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]=A$

Therefore, the result is true for n = 1.

Let the result be true for n = k.

That is,

$P(k): A^{k}=\left[\begin{array}{ll}1+2 k & -4 k \\ k & 1-2 k\end{array}\right], n \in \mathbf{N}$

Now, we prove that the result is true for n = k + 1.

Consider

$A^{k+1}=A^{k} \cdot A$

$=\left[\begin{array}{ll}1+2 k & -4 k \\ k & 1-2 k\end{array}\right]\left[\begin{array}{lr}3 & -4 \\ 1 & -1\end{array}\right]$

$=\left[\begin{array}{ll}3(1+2 k)-4 k & -4(1+2 k)+4 k \\ 3 k+1-2 k & -4 k-1(1-2 k)\end{array}\right]$

$=\left[\begin{array}{ll}3+6 k-4 k & -4-8 k+4 k \\ 3 k+1-2 k & -4 k-1+2 k\end{array}\right]$

$=\left[\begin{array}{ll}3+2 k & -4-4 k \\ 1+k & -1-2 k\end{array}\right]$

$=\left[\begin{array}{ll}1+2(k+1) & -4(k+1) \\ 1+k & 1-2(k+1)\end{array}\right]$

Therefore, the result is true for n = k + 1.

Thus, by the principle of mathematical induction, we have:

$A^{n}=\left[\begin{array}{ll}1+2 n & -4 n \\ n & 1-2 n\end{array}\right], n \in \mathbf{N}$

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