# cofactor expansion

Let $M$ be an $n\times n$ matrix with entries $M_{ij}$ that are elements of a commutative ring. Let $m_{ij}$ denote the determinant   of the $(n-1)\times(n-1)$ submatrix  obtained by deleting row $i$ and column $j$ of $M$, and let

 $C_{ij}=(-1)^{i+j}m_{ij}.$

The subdeterminants $m_{ij}$ are called the minors of $M$, and the $C_{ij}$ are called the cofactors.

We have the following useful formulas for the cofactors of a matrix. First, if we regard $\det M$ as a polynomial in the entries $M_{ij}$, then we may write

 $C_{ij}=\frac{\partial M}{\partial M_{ij}}$ (1)

Second, we may regard the determinant of $M=(M_{1},\ldots,M_{n})$ as a multi-linear, skew-symmetric function  of its columns:

 $\det M=\det(M_{1},\ldots,M_{n}).$

This point of view leads to the following formula:

 $C_{ij}=\det(M_{1},\ldots,\hat{M_{j}},\mathbf{e}_{i},\ldots,M_{n}),$ (2)

where the notation indicates that column $j$ has been replaced by the $i$th standard vector.

As a consequence, we obtain the following representation of the determinant in terms of cofactors:

 $\displaystyle\det(M)$ $\displaystyle=\det(M_{1},\ldots,M_{1j}\mathbf{e}_{1}+\cdots+M_{nj}\mathbf{e}_{% n},\ldots,M_{n})$ $\displaystyle=\sum_{i=1}^{n}M_{ij}C_{ij},\quad j=1,\ldots,n.$

The above identity  is often called the cofactor expansion of the determinant along column $j$. If we regard the determinant as a multi-linear, skew-symmetric function of $n$ row-vectors, then we obtain the analogous cofactor expansion along a row:

 $\displaystyle\det(M)$ $\displaystyle=\sum_{i=1}^{n}M_{ji}C_{ji}.$

## Example.

Consider a general $3\times 3$ determinant

 $\left|\begin{matrix}a_{1}&a_{2}&a_{3}\\ b_{1}&b_{2}&b_{3}\\ c_{1}&c_{2}&c_{3}\end{matrix}\right|=a_{1}b_{2}c_{3}+a_{2}b_{3}c_{1}+a_{3}b_{1% }c_{2}-a_{1}b_{3}c_{2}-a_{3}b_{2}c_{1}-a_{2}b_{1}c_{3}.$

The above can equally well be expressed as a cofactor expansion along the first row:

 $\displaystyle\left|\begin{matrix}a_{1}&a_{2}&a_{3}\\ b_{1}&b_{2}&b_{3}\\ c_{1}&c_{2}&c_{3}\end{matrix}\right|$ $\displaystyle=a_{1}\left|\begin{matrix}b_{2}&b_{3}\\ c_{2}&c_{3}\end{matrix}\right|-a_{2}\left|\begin{matrix}b_{1}&b_{3}\\ c_{1}&c_{3}\end{matrix}\right|+a_{3}\left|\begin{matrix}b_{1}&b_{2}\\ c_{1}&c_{2}\end{matrix}\right|$ $\displaystyle=a_{1}(b_{2}c_{3}-b_{3}c_{2})-a_{2}(b_{1}c_{3}-b_{3}c_{1})+a_{3}(% b_{1}c_{2}-b_{2}c_{1});$

or along the second column:

 $\displaystyle\left|\begin{matrix}a_{1}&a_{2}&a_{3}\\ b_{1}&b_{2}&b_{3}\\ c_{1}&c_{2}&c_{3}\end{matrix}\right|$ $\displaystyle=-a_{2}\left|\begin{matrix}b_{1}&b_{3}\\ c_{1}&c_{3}\end{matrix}\right|+b_{2}\left|\begin{matrix}a_{1}&a_{3}\\ c_{1}&c_{3}\end{matrix}\right|-c_{2}\left|\begin{matrix}a_{1}&a_{3}\\ b_{1}&b_{3}\end{matrix}\right|$ $\displaystyle=-a_{2}(b_{1}c_{3}-b_{3}c_{1})+b_{2}(a_{1}c_{3}-a_{3}c_{1})-c_{2}% (a_{1}b_{3}-a_{3}b_{1});$

or indeed as four other such expansion corresponding to rows 2 and 3, and columns 1 and 3.

Title cofactor expansion CofactorExpansion 2013-03-22 12:01:07 2013-03-22 12:01:07 rmilson (146) rmilson (146) 17 rmilson (146) Theorem msc 15A15 Laplace expansion cofactor minor subdeterminant SarrusRule