Finding the determinant of an $n\times n$ matrix where $det(A)=det(M)\cdot det(N)$. 
Consider an $n\times n$ matrix of the form
$$
A=\begin{pmatrix} M & P \\ 0 & N \\\end{pmatrix}
$$
where $M\in K^{r\times r},P\in K^{r\times s},N\in K^{s\times s}$, and $0\in K^{s\times r}$ is the zero $s\times r$ matrix, $n=r+s$. Show that for such matrices, we have
$$
det(A)=det(M)\cdot det(N).
$$

To answer this question, I think that I want to use row operations on $M$ and $N$ that bring $M, N$ and $A$ into upper triangular form, but I'm not sure on how to do this with "unknown" matrices.
 A: @RobertTheTutor is right here. Remember, for upper triangular matrices, the product of the values on the diagonal equals the determinant. You can reduce any matrix into an upper triangular form. This is because as shown here in method 2, that there always exist a $P$ such that:
$$A_{tri} = P^{-1} * A * P $$
By determinant properties,
$$det(A_{tri}) = det(P^{-1}) * det(A) * det(P) = \frac{1}{det(P)} * det(A) * det(P) = det(N)$$
$$det(A_{tri}) == det(A)$$
You just need ot show that the value of $P$ doesn't affect anything. This works since we don't care what happens to $P$, so just reduce $\begin{bmatrix} M & P\end{bmatrix}$ as if you are reducing $M$ alone, and you will get $M_{tri}$ on the left. Do the same with $\begin{bmatrix}0 & N\end{bmatrix}$, and since there are only zeros on the left, you will still just have zeros on the left. Now you have:
$\begin{bmatrix} M_{tri} & P_{junk} \\
 0 & N_{tri}\end{bmatrix}$
Which is an upper triangular matrix where the product along the diagonal equals the product along $M_{tri}$ times the product along $N_{tri}$, which is $det(A) = det(M)*det(N)$
