How find this $M(2)$ let matrix $$A(x)=\begin{bmatrix}
1&x\\
x&1
\end{bmatrix}$$
and consider the infinte matrix product
$$M(t)=\prod_{n=1}^{\infty}A(p^{-t}_{n})$$
where $p_{n}$ is the nth prime
Evaluate $M(2)$
My try: since
$$M(2)=\prod_{n=1}^{\infty}A(\dfrac{1}{p^2_{n}})=\begin{bmatrix}
1&\dfrac{1}{2^2}\\
\dfrac{1}{2^2}&1
\end{bmatrix}\cdot\begin{bmatrix}
1&\dfrac{1}{3^2}\\
\dfrac{1}{3^2}&1
\end{bmatrix}\cdot\begin{bmatrix}
1&\dfrac{1}{5^2}\\
\dfrac{1}{5^2}&1
\end{bmatrix}\cdots\begin{bmatrix}
1&\dfrac{1}{p_{n}^2}\\
\dfrac{1}{p_{n}^2}&1
\end{bmatrix}\cdots$$
then I can't,Thank you
 A: Your matrices form a commuting family and therefore can be simultaneously diagonalized. Let $P$ be the orthogonal matrix which simultaneously diagonalizes the family. We have
$$P = \begin{pmatrix}-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{pmatrix}$$ 
Note that the eigenvalues of each matrix are $1\pm x$. Then we can write the product as
$$M(t) = P\begin{pmatrix}\prod_{i=1}^\infty\left(1 - \frac{1}{p_i^t}\right) & 0 \\ 0 & \prod_{i=1}^\infty\left(1 + \frac{1}{p_i^t}\right)\end{pmatrix}P^\mathrm{T}$$
The two infinite products are well known as
$$\prod_{i=1}^\infty\left(1 - \frac{1}{p_i^t}\right) = \frac{1}{\zeta(t)}$$
$$\prod_{i=1}^\infty\left(1 + \frac{1}{p_i^t}\right) = \frac{\zeta(t)}{\zeta(2t)}$$
where $\zeta$ is the Riemann zeta. We have $\zeta(2) = \frac{\pi^2}{6}$ and $\zeta(4) = \frac{\pi^4}{90}$ whence we have
$$M(2) = P\begin{pmatrix}\frac{6}{\pi^2} & 0 \\ 0 & \frac{15}{\pi^2}\end{pmatrix}P^\mathrm{T} = \frac{3}{2\pi^2}\begin{pmatrix}7 & 3\\3& 7\end{pmatrix}$$
