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As per comments, I've added the original question below (I'm new here, sorry!): $$A^n=\begin{bmatrix}1&3^n - 1\\0&3^n\end{bmatrix}$$

for all integers n ≥ 1, where $$A = \begin{bmatrix}1&2\\0&3\end{bmatrix}$$

I want to prove that n is true when n = k+1. I have figured out the base step. That leaves the recursive step. I made k+1 of the form

$$k+1 = A*A^n = \begin{bmatrix}1&2\\0&3\end{bmatrix}\begin{bmatrix}1&3^n - 1\\0&3^n\end{bmatrix} $$

Multiplying the 2 matrices I get:

$$k+1 = \begin{bmatrix}1&3^n - 1 + 2*3^n\\0&3*3^n\end{bmatrix} $$

I don't know how to go any further except to simplify the bottom-right term, where it becomes 3^(n+1)

Haneef Mohd Fairuz
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    You can't because they are not equal. If you plug in $n=1$ the first gives $2+2\cdot 3=8$ and the second gives $(9-1)+2(9)=26$ Please check how you got here and correct the problem. – Ross Millikan May 23 '17 at 05:22
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    There is too little to help us here. It is certainly not the case that $(3^n - 1) + 2(3^n) = (3^{(n+1)} - 1) + 2(3^{(n+1)})$. If we are to help you, we will need to know more, in particular what you have tried. – Hans Hüttel May 23 '17 at 05:22
  • $(3^n - 1) + 2 \cdot 3^n=3\cdot 3^n-1=3^{n+1}-1 \ne (3^{n+1}-1) + 2 \cdot 3^{n+1}$ – dxiv May 23 '17 at 05:25

1 Answers1

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The (edit: previously) posted relation is wrong, what the induction proof actually needs is: $$A^{n+1}=A^n \, A = \begin{bmatrix}1&3^n - 1\\0&3^n\end{bmatrix} \, \begin{bmatrix}1&2\\0&3\end{bmatrix} = \begin{bmatrix}1 \cdot 1 + (3^n-1) \cdot 0 & 1 \cdot 2 + (3^n - 1) \cdot 3 \\ 0 \cdot 1 + 3^n \cdot 0 & 0 \cdot 3 + 3 \cdot 3^n\end{bmatrix}$$

For the top-right element, note that $\,2+(3^n-1)\cdot3 = 3 \cdot 3^n + 2 - 3 = 3^{n+1}-1\,$.

dxiv
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  • Have I arranged my multiplication incorrectly? It shouldn't be A^n+1 = A * A^n but A^n * A? – Haneef Mohd Fairuz May 23 '17 at 05:55
  • @HaneefMohdFairuz It doesn't matter, it's the same thing. The way you arranged it now (after the edit), look at my previous comment directly under your question. The end result is the same, either way. – dxiv May 23 '17 at 05:57