What is the cofactor expansion method to finding the determinant?

1 Answer
Feb 18, 2015

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 :

enter image source here

If we develop on the 2nd column, you get

enter image source here

so : enter image source here

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.