We define $\mathbf A = \mathbf K_1 \mathbf K_2 ... \mathbf K_n$ where $\mathbf K_i$ is a sequence of $2 \times 2$ matrices. Using determinant property, the following algebraic equation: $$\det(\mathbf A) = 0$$ can be simplified as: $$\det(\mathbf A) = \prod_{i=1}^n \det(\mathbf K_i) = 0$$ and conclude that: $$\det(\mathbf K_1) = 0 \lor \det(\mathbf K_2) = 0 \lor ... \lor \det(\mathbf K_n) = 0$$
Now consider the case where an Identity matrix is subtracted: $$\det(\mathbf A-\mathbf I)=0$$ I could simplify the equation as follows: $$ \det(\mathbf A - \mathbf I) = \det(\mathbf A) - trace(\mathbf A) + 1=0$$ However, I am wondering that this equation can be further simplified. I think, since the above is similar to $\det(\mathbf A-\lambda \mathbf I)=0$, the characteristic equation of linear eigen-value problems; there has to be something related.