GROUPS OF MATRICES 127
And here is its inverse H:
EXERCISE: Check that AH = HA = I.
It is a fact, easy to prove, that if A is invertible, then it has one
and only one inverse.
This we call the inverse of A, and we write it
. Notice this doesn’t mean
, for A is not a number, so it cannot
be divided into the number 1.
We claimed that not every square matrix has an inverse. Here is
a stupid example. If C is the 3-by-3 matrix of all 0’s, then CB = C
no matter what 3-by-3 matrix B is. So CB could never equal I. But
there are less stupid examples. If you think about it a little, you
can see that even if just one row of A is all 0’s, then A has no
inverse matrix. Or if two of the rows of A are identical, then A has
no inverse matrix. Or if two of the columns of A are identical, then A
has no inverse matrix. Do you see why?
The general theory of which A’s are invertible and which are not
is a very interesting part of matrix theory. The complete answer is
too much off the track for us to go into it here, but if you happen to
know how to compute the determinant of a square matrix, then you
can understand the following:
THEOREM 11.1: If A is a square R-matrix, then A is
invertible if and only its determinant (which is an element
of R) has a multiplicative inverse in R. That is to say, if d is
the determinant of A, then A is invertible if and only if there
is some element c in R so that the product cd = 1.
PROOF:IfH and K are both inverses of A, then H = HI = H(AK) = (HA)K = IK = K.
For example, if A has two identical rows, then so does AB for any matrix B. But the
identity matrix does not have two identical rows. So AB can never equal I, whatever
matrix B you try.