Let $A$ be an $m\times n$ matrix and $B$ an $n\times m$ matrix. Then the determinant of their product $C=AB$ can be written as a sum of products of minors of $A$ and $B$ : \begin{equation*} |C| = \sum_{1\le k_1<k_2<\cdots<k_m\le n} A\begin{pmatrix} 1 & 2 & \cdots & m \\ k_1 & k_2 & \cdots & k_m \end{pmatrix} B\begin{pmatrix} k_1 & k_2 & \cdots & k_m \\ 1 & 2 & \cdots & m \end{pmatrix}. \end{equation*}Basically, the sum is over the maximal ($m$ -th order) minors of $A$ and $B$ . See the entry on minors for notation.

If $m>n$ , then neither $A$ nor $B$ have minors of rank $m$ , so $|C|=0$ . If $m=n$ , this formula reduces to the usual multiplicativity of determinants $|C|=|AB|=|A||B|$ .

Proof. Since $C=AB$ , we can write its elements as $c_{ij} = \sum_{k=1}^n a_{ik} b_{kj}$ . Then its determinant is

In both steps above, we have used the property that the determinant is multilinear in the colums of a matrix.

Note that the terms in the last sum with any two $k$ 's the same will make the minor of $A$ vanish. And, for $\{k_1,\cdots,k_m\}$ 's that differ only by a permutation, the minor of $A$ will simply change sign according to the parity of the permutation. Hence the determinant of $C$ can be rewritten as

where $S_m$ is the permutation group on $m$ elements. But the last sum is none other than the determinant $B \left(\begin{smallmatrix} k_1 & k_2 & \cdots & k_m \\ 1 & 2 & \cdots & m \end{smallmatrix}\right)$ . Hence we write \begin{equation*} |C| = \sum_{1\le k_1<\cdots<k_m\le n} A \begin{pmatrix} 1 & 2 & \cdots & m \\ k_1 & k_2 & \cdots & k_m \end{pmatrix} B \begin{pmatrix} k_1 & k_2 & \cdots & k_m \\ 1 & 2 & \cdots & m\end{pmatrix}, \end{equation*}which is the Cauchy-Binet formula.