When working on the proof for $$\det(\text{A} -\lambda \text{I})=\det(\text{Q}^{-1}\text{ B Q}-\lambda \text{I})$$ where $\lambda$ is a scalar, $\det$ is the determinant, $\text{I}$ is the identity matrix and $\text{A}$ is similar to $\text{B}$. I made the following calculation: $$\det(\text{A} -\lambda \text{I})= \sum^n_{j=1}(a_{kj} - \lambda e_{kj}) \ \text{C}_{kj}$$ where $a_{ij},e_{ij}$ are entries in matricies $\text{A}$ and $\text{I}$ respectively, and $\text{C}_{ij}$ is the co-factor of $(a_{ij} - \lambda e_{ij})$. Now, I cannot distribute the summation function because $\text{C}_{ij}$ relates the the matrix $\text{A} - \lambda \text{I}$ and not the two matricies taken separately. $$\det(\text{A} -\lambda \text{I}) \neq \sum^n_{j=1}a_{kj} \ \text{C}_{kj} - \lambda \sum^n_{j=1}e_{kj} \text{C}_{kj}$$
The proof should be calculated using left and right distributive laws of matrix multiplication, noting that $$\det(\text{Q}^{-1}\text{ B Q}-\lambda \text{I})=\det(\text{Q}^{-1}\text{BQ}-\lambda \text{I Q}^{-1} \text{Q})$$