TITLE: Determinant of a block upper triangular matrix
QUESTION [27 upvotes]: How prove the following equality for a block matrix?
$$\det\left[\begin{array}[cc]\\A&C\\
0&B\end{array}\right]=\det(A)\det(B)$$
I tried to use a proof by induction but I'm stuck. Is there a simpler method? Thanks for help.

REPLY [16 votes]: Since I have criticised some other answers, let me be fair and give my take on this. The result to use is just the Leibniz formula defining the determinant (for once, use the definition!):
$$
\def\sg{\operatorname{sgn}}
\det(M) = \sum_{\sigma \in S_n} \sg(\sigma) \prod_{i = 1}^n M_{i,\sigma(i)}.
$$
Now if $M$ is the matrix of the question, and its block $A$ has size $k\times k$, then by the block form $M_{i,j}=0$ whenever $j\leq k<i$ (lower left hand block). So we can drop all permutations from the sum for which $\sigma(i)\leq k$ for any $i\in\{k+1,\ldots,n\}$. But that means that for $\sigma$ to survive, the $n-k$ values $\sigma(i)$ for are $i\in T=\{k+1,\ldots,n\}$ must all be in the $n-k$-element set $T$, and they must of course also be distinct: they form a permutation of the elements of$~T$. So $\sigma$ permutes them elements of $T$ among each other, and then necessarily also the elements of the complement $\{1,\ldots,k\}$ of$~T$. These permutations of subsets are independent, so the surviving permutations are naturally in bijection with the Cartesian product of the permutations of $\{1,\ldots,k\}$ and those of$~T$. Also the sign of the combination of two such permutations is the product of the signs of the individual permutations. All in all one gets
$$
\begin{align}
  \det(M)&=\sum_{\pi \in S_k} \sum_{\tau\in S_T} \sg(\pi)\sg(\tau) \left(\prod_{i = 1}^k M_{i,\pi(i)}\right)\prod_{i \in T}M_{i,\tau(i)}
\\&=\left(\sum_{\pi \in S_k}\sg(\pi)\prod_{i = 1}^k M_{i,\pi(i)}\right)\left(\sum_{\tau\in S_T}\sg(\tau)\prod_{i \in T}M_{i,\tau(i)}\right)
\\&=\det (A)\det(B).
\end{align}
$$