For any matrices $A$,$B$ in $SL(2,\mathbb{C})$, let $M(A,B)$ be the 4 by 4 matrix whose columns are matrices $I$,$A$,$B$,$AB$. Then it is not hard to verify that $det M(A,B)= 2 tr[A,B]$. Is there an analogous identity for matrices in $SL(3,\mathbb{C})$? I suppose if there is one, then $M(A,B)$ would be $9$ by $9$ matrix whose columns are matrices $I$,$A$,$A^{2}$,$B$,$AB$,$A^{2}B$, $B^{2}$, $AB^{2}$, $A^{2}B^{2}$.

Given $X_1,\dots,X_{n^2} \in M_n(\mathbb{C})$, one can form the $n^2 \times n^2$ matrix whose $i$th column is the entries of $X_i$ in some fixed order. The determinant of this matrix (defined up to sign) is called the discriminant of $X_i, \dots , X_{n^2}$ and is denoted $\mathcal{D}(X_i)$. It is a matrix invariant, which means that $\mathcal{D}(X_i) = \mathcal{D}(UX_iU^{1})$ for any $U \in GL(n ,\mathbb{C})$. The first fundamental theorem of matrix invariants says that any matrix invariant can be expressed in terms of traces. The expression is not unique, since there are "trace identities" which are identically zero on $M_n(\mathbb{C})$. An explicit formula for $\mathcal{D}$ in terms of traces is given on p.46 of my book "The Polynomial Identities and Invariants of $n\times n$ Matrices". It happens that for $2 \times 2$ matrices, $\mathcal{D}(I,A,B,AB)I = \pm(AB  BA)^2$. In other words, $\mathcal{D}(I,A,B,AB)$ is the value of a central polynomial for $2 \times 2$ matrices. As far as I know, it is not known if $\mathcal{D}(A^iB^j \mid 0 \leq i,j \leq n1)$ is the value of a central polynomial for $n \geq 3$. 

