If you have a (numerical) Grothendieck group, it comes equipped with an Euler form and usually it has a Serre automorphism too. If the numerical Grothendieck group happens to be finitely generated (i.e. isomorphic to $\mathbb{Z}^n$ for some $n$), then choosing a basis you can express the Euler form (which is a bilinear form) and the Serre automorphism as a matrix in terms of this basis. In that case the terminology that is used in the literature is:

Euler form $\langle-,-\rangle$
Gram matrix
Cartan matrix
Serre automorphism $\mathbb{S}$
Coxeter matrix

The Serre automorphism is sometimes called Coxeter transformation too.

If we denote the Cartan–Gram matrix for the Euler form by $A$ and the Coxeter matrix for the Serre automorphism by $C$, then there is a strong relationship between the two.

Proposition We have that $$C=-A^{-1}A^{\mathrm{t}}.$$

Proof. Because $\langle-,-\rangle$ is nondegenerate, we have $\det A\neq 0$, and \begin{aligned} \langle y,x\rangle &=y^{\mathrm{t}}Ax \\ &=x^{\mathrm{t}}A^{\mathrm{t}}y \\ &=-x^{\mathrm{t}}A(-A^{-1}A^{\mathrm{t}})y \\ &=\left\langle x,(-A^{-1}A^{\mathrm{t}})y \right\rangle \end{aligned} so we get the desired relationship by nondegeneracy. $\blacksquare$

To finish this blogpost, let's give the two easiest examples: curves and quivers.

Example 1 The numerical Grothendieck group of a smooth projective curve $C$ is generated by $[k(p)]$ and $[\mathcal{O}_C]$. You can easily compute using Riemann–Roch that $$A= \begin{pmatrix} 0 & -1 \\ 1 & 1-g \end{pmatrix}$$ and $$C= \begin{pmatrix} 1 & 2g-2 \\ 0 & 1 \end{pmatrix},$$ where $g$ is the genus of the curve.

Example 2 The (numerical) Grothendieck group of a path algebra for an acyclic quiver has two obvious choices of basis: the simples, and the projectives. We will use the projectives here. Just note there is a strong connection between the Serre functor and Auslander–Reiten triangles, see e.g. theorem A of Reiten–Van den Bergh, Noetherian hereditary abelian categories.

If we take the Dynkin quiver $\mathrm{D}_4$ as an example, and we orient the arrows from left to right, then $$A= \begin{pmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$ and $$C= \begin{pmatrix} 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & -1 & -1 & -1 \end{pmatrix}.$$