
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
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 $\mathrm{det}M$ is on the lefthand side in (A) instead of 1, and ${(1)}^{i+j}{M}_{ij}$ appears in (A) and (B) instead of ${M}_{ji}^{1}$ here.
We can therefore divide both sides of (A) and (B) by $\mathrm{det}M$ , 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 ${M}_{ji}^{1}$ 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).)
