An explicit representation of a free nilpotent group by unitriangular matrices P. Hall proved that every finitely generated torsion-free nilpotent group can be faithfully represented by upper unitriangular matrices over $\mathbb Z$. The most famous example is the integral Heisenberg group, which is free nilpotent, of rank $2$ and class $2$, and is isomorphic to $UT(3,\mathbb{Z})$.
I would like to know what is the easiest such representation for the free nilpotent group $F(3,2)$, of rank $3$ and class $2$. Its group presentation is
$$F(3,2)=\langle x,y,z \,\|\, [x,[x,y]],[y,[x,y]],[z,[x,y]],[x,[y,z]],[y,[y,z]],[z,[y,z]],[x,[x,z]],[y,[x,z]],[z,[x,z]] \rangle.$$
An algorithm to compute this representation was implemented in GAP, based on the work of de Graaf and Nickel [de Graaf, Willem A.; Nickel, Werner, Constructing faithful representations of finitely-generated torsion-free nilpotent groups, J. Symb. Comput. 33, No. 1, 31-41 (2002). ZBL1021.20006.] Unfortunately I am not proficient with GAP to run it myself.
In the above paper the authors state that the algorithm produces a representation for $F(3,2)$ in $UT(6,\mathbb{Z})$.
Is there a faithful representation of $F(3,2)$, over integer matrices, of degree less than $6$?
 A: I think that I have found an embedding of $F(3,2)$ into ${\rm UT}(5,{\mathbb Z})$. I used a combination of GAP and Magma to do this, but here is the (putative) result in GAP.
The images of the three generators of $F(3,2)$ are
a := [
    [ 1, 0, 1, 0, -1 ],
    [ 0, 1, 0, 0, 0 ],
    [ 0, 0, 1, 0, -1 ],
    [ 0, 0, 0, 1, 0 ],
    [ 0, 0, 0, 0, 1 ]
];;
b := [
    [ 1, 0, 0, 0, 0 ],
    [ 0, 1, 0, 1, 0 ],
    [ 0, 0, 1, 0, -1 ],
    [ 0, 0, 0, 1, 0 ],
    [ 0, 0, 0, 0, 1 ]
];;
c := [
    [ 1, 0, 0, 0, 0 ],
    [ 0, 1, 1, 1, -1 ],
    [ 0, 0, 1, 1, -1 ],
    [ 0, 0, 0, 1, 0 ],
    [ 0, 0, 0, 0, 1 ]
];;

If you calculate the three commutators $[a,b]$, $[a,c]$ and $[b,c]$, you will find that they clearly generate a subgroup isomorphic to ${\mathbb Z}^3$, and they are all centralized by $a$,$b$, and $c$, which I think is enough to show that the group they generate is isomorphic to $F(3,2)$.
gap> x:=Comm(a,b);; y := Comm(a,c);; z := Comm(b,c);;
gap> Display(x); Display(y); Display(z);
[ [   1,   0,   0,   0,  -1 ],
  [   0,   1,   0,   0,   0 ],
  [   0,   0,   1,   0,   0 ],
  [   0,   0,   0,   1,   0 ],
  [   0,   0,   0,   0,   1 ] ]
[ [   1,   0,   0,   1,  -1 ],
  [   0,   1,   0,   0,   1 ],
  [   0,   0,   1,   0,   0 ],
  [   0,   0,   0,   1,   0 ],
  [   0,   0,   0,   0,   1 ] ]
[ [  1,  0,  0,  0,  0 ],
  [  0,  1,  0,  0,  1 ],
  [  0,  0,  1,  0,  0 ],
  [  0,  0,  0,  1,  0 ],
  [  0,  0,  0,  0,  1 ] ]

A: The group $F(r,c)$ embeds in the $\mathbf{Q}$-group associated to the free $c$-step nilpotent Lie $\mathbf{Q}$-algebra on $r$ generators.
The latter possesses a natural grading in $\{1,\dots,r\}$.
More generally if a finite-dimensional Lie algebra $g$ over a field $K$ is graded in $\mathbf{Z}$ with $g_0=\{0\}$ [this can be shown to imply $g$ nilpotent], then it has a natural faithful representation in dimension $\dim(g)+1$.
Indeed, first define $D$ as the (invertible) derivation of $g$ acting as $\times k$ on $g_k$. Let $h$ be the semidirect product $K\ltimes g$, where $1\in K$ acts as $D$. Then $h$ has trivial center, so its adjoint representation is faithful, hence is faithful in restriction to $g$. With a little effort and exponentiation one should be able to provide explicit matrices for the corresponding representation of $F(r,c)$.
This is the easiest faithful representation I can figure out, which doesn't mean it's the easiest. And doesn't mean it's the smallest one (indeed for $F(3,2)$ this yields dimension 7, while other answers provide a 5-dimensional one — for $F(r,2)$ it yields dimension $1+r+r(r-1)/2$).
