This is probably well-known to representation theorists, but this doesn't imply being well-known to me.

Let $k$ be a field, and let $A$ be a $k$-algebra that is finite-dimensional as a $k$-vector space. Let $S_1, S_2, \ldots, S_m$ be a complete list of pairwise non-isomorphic simple $A$-modules (up to isomorphism). Let $P_1, P_2, \ldots, P_m$ be the projective covers of $S_1, S_2, \ldots, S_m$, respectively. (Thus, $P_1, P_2, \ldots, P_m$ is a complete list of pairwise non-isomorphic indecomposable projective $A$-modules.) The *Cartan matrix* of $A$ is defined to be the $m\times m$-matrix in $\mathbb{Z}^{m\times m}$ whose $\left(i,j\right)$-th entry (for all $i$ and $j$) is the number of composition factors of $P_j$ isomorphic to $S_i$. (In other words, its $\left(i,j\right)$-th entry is $\dfrac{\dim \operatorname{Hom}_A \left(P_i, P_j\right)}{\dim \operatorname{End}_A \left(S_i\right)}$.) See §7.4 of Peter Webb's *A Course in Finite Group Representation Theory* for proofs and some background.

**1.** Is it true that the Cartan matrix of $A$ is invertible as a matrix in $\mathbb{Q}^{m\times m}$ ? (I think the answer is "no", but I don't know of a counterexample.)

**2.** Does this change if we require $k$ to be algebraically closed (or at least to be a splitting field for $A$) ?

**3.** Does this change if we furthermore require $A$ to be a Hopf algebra?

What I know is that the Cartan matrix of $A$ is invertible if $A$ is the group algebra of a finite group. This is proven in Corollary 10.2.4 of Peter Webb's above-mentioned book, but the proof uses the theory of Brauer characters, which as far as I know is particular to group algebras. (Or not? Is there a notion of Brauer characters for Hopf algebras? I certainly wouldn't find it strange, at least compared with the strangeness of classical Brauer character theory, but I have never seen such a notion.)