Editing Cauchy–Binet formula (section)

== Special cases ==
If ''n''&nbsp;&lt;&nbsp;''m'' then <math>\tbinom{[n]}m</math> 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 (linear algebra)|rank]] of the ''m''×''m'' 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, <math>\tbinom{[n]}m=\{[n]\} </math> (a [[singleton (mathematics)|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 matrix|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;Ø, and the formula states 1&nbsp;=&nbsp;1, with both sides given by the determinant of the 0×0 matrix. For ''m''&nbsp;=&nbsp;1, the summation ranges over the collection <math>\tbinom{[n]}1</math> of the ''n'' different singletons taken from [''n''], and both sides of the formula give <math>\textstyle\sum_{j=1}^nA_{1,j}B_{j,1}</math>, the [[dot product]] of the pair of [[Tuple|vector]]s represented by the matrices. The smallest value of ''m'' for which the formula states a non-trivial equality is ''m''&nbsp;=&nbsp;2; it is discussed in the article on the [[Binet–Cauchy identity]].

=== In the case ''n''&nbsp;=&nbsp;3 ===
Let <math> \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}, \boldsymbol{d}, \boldsymbol{x}, \boldsymbol{y},\boldsymbol{z},\boldsymbol{w} </math> be three-dimensional vectors.

: <math>
\begin{align}
& 1 = 1 & (m = 0)\\[10pt]
& \boldsymbol{a}\cdot \boldsymbol{x} =  a_1  x_1 + a_2 x_2 + a_3 x_3 & (m = 1)\\[10pt]
& \begin{vmatrix}
\boldsymbol{a}\cdot \boldsymbol{x} & \boldsymbol{a}\cdot \boldsymbol{y}\\
\boldsymbol{b}\cdot \boldsymbol{x} & \boldsymbol{b}\cdot \boldsymbol{y}
\end{vmatrix} \\[4pt]
= {} &  
\begin{vmatrix}
a_2 & a_3\\
b_2 & b_3
\end{vmatrix}
\begin{vmatrix}
x_2 & y_2\\
x_3 & y_3
\end{vmatrix}
+
\begin{vmatrix}
a_3 & a_1\\
b_3 & b_1
\end{vmatrix}
\begin{vmatrix}
x_3 & y_3\\
x_1 & y_1
\end{vmatrix}
+
\begin{vmatrix}
a_1 & a_2\\
b_1 & b_2
\end{vmatrix}
\begin{vmatrix}
x_1 & y_1\\
x_2 & y_2
\end{vmatrix}\\[4pt]
= {} & (\boldsymbol{a}\times\boldsymbol{b})\cdot(\boldsymbol{x}\times\boldsymbol{y})
& (m = 2)\\[10pt]
& \begin{vmatrix}
\boldsymbol{a}\cdot \boldsymbol{x} & \boldsymbol{a}\cdot \boldsymbol{y} & \boldsymbol{a}\cdot \boldsymbol{z}\\ 
\boldsymbol{b}\cdot \boldsymbol{x} & \boldsymbol{b}\cdot \boldsymbol{y} & \boldsymbol{b}\cdot \boldsymbol{z}\\ 
\boldsymbol{c}\cdot \boldsymbol{x} & \boldsymbol{c}\cdot \boldsymbol{y} & \boldsymbol{c}\cdot \boldsymbol{z} 
\end{vmatrix}
=  
\begin{vmatrix}
a_1 & a_2 & a_3\\
b_1 & b_2 & b_3\\
c_1 & c_2 & c_3
\end{vmatrix}
\begin{vmatrix}
x_1 & y_1 & z_1 \\
x_2 & y_2 & z_2 \\
x_3 & y_3 & z_3
\end{vmatrix}\\[4pt]
= {} & [\boldsymbol{a}\cdot (\boldsymbol{b} \times \boldsymbol{c})] [\boldsymbol{x}\cdot (\boldsymbol{y} \times \boldsymbol{z})] 
& (m = 3)\\[10pt]
& \begin{vmatrix}
\boldsymbol{a}\cdot \boldsymbol{x} & \boldsymbol{a}\cdot \boldsymbol{y} & \boldsymbol{a}\cdot \boldsymbol{z} & \boldsymbol{a}\cdot \boldsymbol{w} \\ 
\boldsymbol{b}\cdot \boldsymbol{x} & \boldsymbol{b}\cdot \boldsymbol{y} & \boldsymbol{b}\cdot \boldsymbol{z} & \boldsymbol{b}\cdot \boldsymbol{w} \\ 
\boldsymbol{c}\cdot \boldsymbol{x} & \boldsymbol{c}\cdot \boldsymbol{y} & \boldsymbol{c}\cdot \boldsymbol{z} &
\boldsymbol{c}\cdot \boldsymbol{w} \\ 
\boldsymbol{d}\cdot \boldsymbol{x} & \boldsymbol{d}\cdot \boldsymbol{y} & \boldsymbol{d}\cdot \boldsymbol{z} &
\boldsymbol{d}\cdot \boldsymbol{w} 
\end{vmatrix}
= 0  & (m = 4)
\end{align}
</math>

In the case ''m''&nbsp;>&nbsp;3, the right-hand side always equals&nbsp;0.