Prove that if M is a n x n matrix that can be written in $\displaystyle M = \left({\begin{array}{cc} A & B \\ 0 & C \end{array}} \right) $
where A and C are square matrices, then det(M) = det(A) det(C)
Write the matrix $\displaystyle A$ in Jordan normal form with $\displaystyle \lambda_1, ..., \lambda_k$ on the diagonal and 0's below the diagonal. Then $\displaystyle \text{det}(A) = \lambda_1 \lambda_2 ...\lambda_k. $ Likewise matrix $\displaystyle C$ can be written in this form and we know $\displaystyle \text{det}(C) = \lambda_{k+1}\lambda_{k+2}...\lambda_n $.
But then the entire matrix $\displaystyle M$ is in Jordan normal form with $\displaystyle \lambda_1, ..., \lambda_n$ on the diagonal and 0's below the diagonal, thus $\displaystyle \text{det}(M) = \lambda_{1}\lambda_{2}...\lambda_n = \text{det}(A)\text{det}(C)$.