
The inverse of a square matrix M is a matrix, denoted as M^{1}, with the property that M^{1} M = M M^{1} = I. Here I is the identity matrix of the same size as M, having 1's on the diagonal and 0's elsewhere.
In terms of transformations, M^{1} undoes the transformation produced by M and so the combination M^{1}M represents the transformation that changes nothing.
The condition MM^{1} = I can be written as
and
when k and i are different, and these conditions completely determine the matrix M^{1} given M, when M has an inverse.
These equations have the same form as the two conditions (A) and (B) of section 4.3 except that det M is on the lefthand side in (A) instead of 1, and (1)^{i + j}M_{ij} appears in (A) and (B) instead of M^{1}_{ji} here.
We can therefore divide both sides of (A) and (B) by det M, and deduce
Remember that here M_{ij} is the determinant of the matrix obtained by omitting the ith row and jth column of M; the elements of M are the m_{ij}, while M^{1}_{ji} here represents the element of the inverse matrix to M in jth row and ith column.
We can phrase this in words as: the inverse of a matrix M is the matrix of its cofactors, with rows and columns interchanged, divided by its determinant.
Exercises:
4.7 Compute the inverse of the matrix in Exercise 4.4 using this formula. Check the product M^{1}M to be sure your result is correct.
4.8 Set up a spreadsheet that computes the inverse of any three by three
matrix with nonzero determinant, using this formula.
(Hint: by copying the first two rows into a fourth and fifth row and the first
two columns into a fourth and fifth column, you can make one entry and copy
to get all of the (1)^{i + j}M_{ij} at once. Then all that
is left is rearranging to swap indices and dividing by the determinant (which
is the dot product of any row of M with the corresponding cofactors).)
