Nothing to do with the determinant. In dimension 2 you can reduce if the matrix is not diagonal--obvious. What about the diagonal case? It is also possible unless a matrix is scalar, i.e. proportional to identity. To do this reduction, we make first a linear transform, so that after it the matrix is not diagonal anymore, and then reduce.
General criteria: System with constant coefficients could be reduced to a single equation iff each eigenspace is $1$-dimensional. To understand why we need to consider solutions to a homogeneous equation and to a homogeneous system.
For an equation one of the solutions is $t^{m-1} e^{kt}$ where $k$ is characteristic root, and $m$ is a multiplicity of $k$.
For system all solutions are in the form $P_{s-1}(t)e^{kt}$ where $P_{s-1}(t)$ are polynomials of degree $\le s-1$ with vector-coefficients and $s$ is the maximal dimension of the corresponding Jordan cells.
Therefore reduction can be done iff $s=m$ which means that for each eigenvalue $k$ there is just one cell, which in turn means, that there is only one linearly independent eigenvector.
Remark: If $s_1,...,s_j$ are dimensions of all cells, corresponding to $k$, then their sum $=m$ where $m$ is a multiplicity of $k$ as a root of characteristic equation, and also the dimension of the root subspace, and $j$ is a dimension of the corresponding eigenspace.
However, it is not important: we solve systems without reducing them to single equations.
