Cryptography assignment question: matrix $A$ is \begin{equation} A = \left(\begin{array}{ccc} 1 & 0 & 0 \\ 1 & 3 & 1 \\ 0 & 2 & 5 \end{array}\right) \end{equation} (Hill Cipher key). Show that for a vector $x_1$, $x_1A=x_2A \mod 26$ for exactly $13$ $x_2$'s.
Progress so far: $\det(A)=13$ so $A$ is not invertible $\mod 26$. Re-write the problem as find $y$ such that $yA=0 \mod 26$. I thought I could find $y$ by finding the null space of $A$ $\mod 26$ but I don't know how to do that. It doesn't seem to be working.