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-1M represents the transformation that changes nothing.
The condition MM-1 = I can be written as
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 left-hand side in (A) instead of 1, and (-1)i + jMij appears in (A) and (B) instead of M-1ji here.
We can therefore divide both sides of (A) and (B) by det M, and deduce
Remember that here Mij is the determinant of the matrix obtained by omitting the i-th row and j-th column of M; the elements of M are the mij, while M-1ji here represents the element of the inverse matrix to M in j-th row and i-th 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.
4.7 Compute the inverse of the matrix in Exercise 4.4 using this formula. Check the product M-1M to be sure your result is correct.
4.8 Set up a spreadsheet that computes the inverse of any three by three
matrix with non-zero 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 + jMij 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).)