# What is the cofactor expansion method to finding the determinant?

Feb 18, 2015

Hello !

Let $A = \left({a}_{i , j}\right)$ be a matrix of size $n \setminus \times n$.
Choose a column : the column number ${j}_{0}$ (I'll write : "the ${j}_{0}$-th column").

The cofactor expansion formula (or Laplace's formula) for the ${j}_{0}$-th column is

$\setminus \det \left(A\right) = \setminus {\sum}_{i = 1}^{n} {a}_{i , {j}_{0}} {\left(- 1\right)}^{i + {j}_{0}} \setminus {\Delta}_{i , {j}_{0}}$

where $\setminus {\Delta}_{i , {j}_{0}}$ is the determinant of the matrix $A$ without its $i$-th line and its ${j}_{0}$-th column ; so, $\setminus {\Delta}_{i , {j}_{0}}$ is a determinant of size $\left(n - 1\right) \setminus \times \left(n - 1\right)$.

Note that the number ${\left(- 1\right)}^{i + {j}_{0}} \setminus {\Delta}_{i , {j}_{0}}$ is called cofactor of place $\left(i , {j}_{0}\right)$.

Maybe it looks like complicated, but it's easy to understand with an example. We want calculate $D$ :

If we develop on the 2nd column, you get

so :

Finally, $D = 0$.

To be efficient, you have to choose a line which has a lot of zeros : the sum will be very simple to calculate !

Remark. Because $\setminus \det \left(A\right) = \setminus \det \left({A}^{\setminus} \textrm{T}\right)$, you can also choose a line rather a column. So, the formula becomes

$\setminus \det \left(A\right) = \setminus {\sum}_{j = 1}^{n} {a}_{{i}_{0} , j} {\left(- 1\right)}^{{i}_{0} + j} \setminus {\Delta}_{{i}_{0} , j}$

where ${i}_{0}$ is the number of the selected line.