Torsion of $\mathbb Z ^{2 \times 2}$ I am trying to find the torsion of the ring $R= \mathbb Z ^{2 \times 2}$ seen as an $R-$ left module. So I want to see for which matrices $B$, there exists $A \neq 0$ such that $AB=0$. So if $B=\begin{pmatrix} a & b \\  c & d  \\  \end{pmatrix}$, then I want to find $A=\begin{pmatrix} x & y \\  z & w  \\  \end{pmatrix}$ with $$\begin{pmatrix} x & y \\  z & w  \\  \end{pmatrix}\begin{pmatrix} a & b \\  c & d  \\  \end{pmatrix}=\begin{pmatrix} 0 & 0 \\  0 & 0  \\  \end{pmatrix}$$
From this condition I get these equations:
$$ax+cy=0$$$$bx+dy=0$$$$az+cw=0$$$$bz+dw=0$$
If $B$ is an invertible matrix in $R$, then it is clear that $B$ is not in the torsion. This led me to think that the torsion may be the set of non invertible matrices, or at least is contained in it. I've tried to find the invertible matrices in this ring but I got stuck.
Any suggestions would be appreciated.

we go ahead and find an inverteble rational matrix P as a multiple of elementary matrices reducing A reduced rational row-echelon form B: PAx=0. Then we solve for x and find its general integral form. From these we may have to discard eigen vectors of P.
