Let $F=\langle x_1,x_2\rangle$ be a free group of rank $2$ and $\Phi=F/F''=\langle \overline{x}_1, \overline{x}_2\rangle$ where $F''$ is second derived subgroup of $F$ (i.e. $F'=[F,F]$ and $F''=[F',F']$).

Let $M=\mathbb{Z}[s_i,s_i^{-1},t_i,t_i^{-1}]_{i=1,2}$, the polynomial ring in eight (commuting) variables over $\mathbb{Z}$ with $s_i.s_i^{-1}=t_i.t_i^{-1}=1$. By universal property (definition) of $F$, the map

$$\rho\colon x_i\mapsto \begin{bmatrix} s_i & t_i\\ 0 & 1\end{bmatrix}$$ extends to a homomorphism from $F$ to ${\rm GL}(M)$, and it can be shown that the subgroup $F''$ of $F$ is in the kernel of homomorphism. Thus we get a homomorphism $$\overline{\rho}\colon F/F''\rightarrow {\rm GL}(M)\,\,\,\, \mbox{ i.e. }\,\,\,\, \overline{\rho}\colon \Phi\rightarrow {\rm GL}(M).$$ Question: Whay $\overline{\rho}$ is faithful? (In other words, why $\ker\rho$ is exactly $F''$?)

$\overline{\rho}$ is called the Magnus representation of the free metabelian group $\Phi$. In the paper

"Automorphisms of Free Matabelian Groups"-Bachmuth,

is was pointed that $\overline{\rho}$ is faithful representation of $\Phi$. But how to prove it? I didn't understand the faithfulness. If this is elementary, then please, give at least a hint.