Question

What is the cofactor expansion method to finding the determinant?

Answer 1

Hello !

Let `A = (a_{i,j})` be a matrix of size `n\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

`\det(A) = \sum_{i=1}^n a_{i,j_0} (-1)^{i+j_0}\Delta_{i,j_0}`

where `\Delta_{i,j_0}` is the determinant of the matrix `A` without its `i`-th line and its `j_0`-th column ; so, `\Delta_{i,j_0}` is a determinant of size `(n-1)\times (n-1)`.

Note that the number `(-1)^{i+j_0}\Delta_{i,j_0}` is called cofactor of place `(i,j_0)`.

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 `\det(A) = \det(A^\text{T})`, you can also choose a line rather a column. So, the formula becomes

`\det(A) = \sum_{j=1}^n a_{i_0,j} (-1)^{i_0+j}\Delta_{i_0,j}`

where `i_0` is the number of the selected line.