in linear algebra, the cauchyâbinet formula, named after augustin-louis cauchy and jacques philippe marie binet, is an identity for the determinant of the product of two rectangular matrices of transpose shapes (so that the product is well-defined and square). it generalizes the statement that the determinant of a product of square matrices is equal to the product of their determinants. the formula is valid for matrices with entries from any commutative ring. == statement == let a be an m&times;n matrix and b an n&times;m matrix. write [n] for the set {&nbsp;1,&nbsp;...,&nbsp;n&nbsp;}, and \tbinom{[n]}m for the set of m-combinations of [n] (i.e., subsets of size m; there are \tbinom nm of them). for s\in\tbinom{[n]}m, write a[m],s for the m&times;m matrix whose columns are the columns of a at indices from s, and b's,[m] for the m&times;m matrix whose rows are the rows of b at indices from s. the cauchyâbinet formula then states : \det(ab) = \sum_{s\in\tbinom{[n]}m} \det(a_{[m],s})\det(b_{s,[m]}). example: taking m&nbsp;=&nbsp;2 and n&nbsp;=&nbsp;3, and matrices a = \begin{pmatrix}1&1&2\\ 3& 1& -1\\ \end{pmatrix} and b = \begin{pmatrix}1&1\\3&1\\0&2\end{pmatrix}, the cauchyâbinet formula gives the determinant: : \det(ab)= \left|\begin{matrix}1&1\\3&1\end{matrix}\right| \cdot \left|\begin{matrix}1&1\\3&1\end{matrix}\right| + \left|\begin{matrix}1&2\\1&-1\end{matrix}\right| \cdot \left|\begin{matrix}3&1\\0&2\end{matrix}\right| + \left|\begin{matrix}1&2\\3&-1\end{matrix}\right| \cdot \left|\begin{matrix}1&1\\0&2\end{matrix}\right|. indeed ab =\begin{pmatrix}4&6\\6&2\end{pmatrix}, and its determinant is â28, which is also the value -2\times-2+-3\times6+-7\times2 given by the right hand side of the formula. == special cases == if n&nbsp;&lt;&nbsp;m then \tbinom{[n]}m is the empty set, and the formula says that det(ab)&nbsp;=&nbsp;0 (its right hand side is an empty sum); indeed in this case the rank of the mÃm matrix ab is at&nbsp;most&nbsp;n, which implies that its determinant is zero. if n = m, the case where a and b are square matrices, \tbinom{[n]}m=\{[n]\} (a singleton set), so the sum only involves s&nbsp;=&nbsp;[n], and the formula states that det(ab)&nbsp;=&nbsp;det(a)det(b). for m&nbsp;=&nbsp;0, a and b are empty matrices (but of different shapes if n&nbsp;&gt;&nbsp;0), as is their product ab; the summation involves a single term s&nbsp;=&nbsp;Ã, and the formula sta
Determinants