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4.4 The Determinant and the Inverse of a Matrix

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 M M − 1 = I can be written as

1 = ∑ j m i j M j i - 1

and

0 = ∑ j m k j M j i - 1

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 + j M i j appears in (A) and (B) instead of M j i − 1 here.

We can therefore divide both sides of (A) and (B) by det ⁡ M , and deduce

M j i − 1 = ( − 1 ) i + j M i j det ⁡ M

Remember that here M i j 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 m i j , while M j i − 1 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.

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 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 + j M i j 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).)

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