Given a matrix $A(x)$, prove that $A(1)*A(2)*...*A(2017)=A(1009^2)$

Question

I am given the matrix $A=\begin{bmatrix}1&x&0\\0&1&0\\0&0&30^x\end{bmatrix}$ and I have to prove:

$A(1)*A(2)*...*A(2017)=A(1009^2)$

From previous sub-points of the problem I have proved that $A(x)*A(y)=A(x+y)$ (1) for any x,y from $R$. Also, I have proved (with mathematical induction) that $A^k=\begin{bmatrix}1&k*x&0\\0&1&0\\0&0&30^{k*x}\end{bmatrix}$ (2),

but I can't see how to use these to prove that $A(1)*A(2)*...*A(2017)=A(1009^2)$.

If I use (1), wouldn't $A(1)*A(2)*...*A(2017)=A(1+2+3+...+2017)$ ? Making the answer $A(2017*1009)$ ? Or am I using the proof (1) wrongly? How should I approach this problem?

Answer 1

It seems that there is a mistake in the statement, however, it is easy to see that \begin{align*} A(1)\cdot A(3) \cdot A(5) \cdot \ldots \cdot A(2017)&=\prod_{k=1}^{1009}A(2k-1)\\[6pt] &=\begin{bmatrix}1&\sum_{k=1}^{1009}(2k-1)&0\\0&1&0\\0&0&30^{\sum_{k=1}^{1009}(2k-1)}\end{bmatrix}\\[6pt] &=\begin{bmatrix}1&1009^2&0\\0&1&0\\0&0&30^{1009^2}\end{bmatrix} \end{align*}

Answer 2

From what you said, I would think there is a mistake in the formulation of the problem. Indeed, your method is correct and you can prove easily that $k\mapsto A(k)$ is injective : $$ \forall k,k'\in \mathbb N, \quad A(k)= A(k') \Rightarrow k= k'.$$ So from what you proved : $$\prod_{k=1}^{2017} A(k) = A\left(\sum_{k=1}^{2017} k\right) = A\left(\frac{2017*2018}{2}\right) = A(2017*1009)$$ If we had $$\prod_{k=1}^{2017} A(k) = A(1009^2),$$ we thus would get : $$ A(1009*1009) = A(2017*1009)$$ and by injectivity of $k\mapsto A(k)$, $$ 1009*1009 = 2017*1009 $$ thus $$ 1009 = 2017 $$ absurd.