2.1.4 Finding The Inverse Of A Matrix
A little later a formula is given for the inverse of a matrix. However, it is not a good way to find
the inverse for a matrix. There is a much easier way and it is this which is presented here. It is
also important to note that not all matrices have inverses.
Example 2.1.24 Let A =
. Does A have an inverse?
One might think A would have an inverse because it does not equal zero. However,
and if A−1 existed, this could not happen because you could multiply on the left by the inverse A
and conclude the vector
. Thus the answer is that A
does not have an
Suppose you want to find B such that AB = I. Let
Also the ith column of I is
Thus, if AB = I, bi, the ith column of B must satisfy the equation Abi = ei. The augmented
matrix for finding bi is
Thus, by doing row operations till A
you end up with
is the solution to Abi
. Now the same sequence of row operations works
regardless of the right side of the agumented matrix
and so you can save trouble by simply
doing the following.
and the ith column of B is bi, the solution to Abi = ei. Thus AB = I.
This is the reason for the following simple procedure for finding the inverse of a matrix. This
procedure is called the Gauss Jordan procedure. It produces the inverse if the matrix has one.
Actually, it produces the right inverse.
Procedure 2.1.25 Suppose A is an n×n matrix. To find A−1 if it exists, form the augmented
n × 2n matrix,
and then do row operations until you obtain an n × 2n matrix of the form
if possible. When this has been done, B = A−1. The matrix A has an inverse exactly when it is
possible to do row operations and end up with one like 2.18.
As described above, the following is a description of what you have just done.
where those Ri
sympolize row operations. It follows that you could undo what you did by doing
the inverse of these row operations in the opposite order. Thus
is the row operation which undoes the row operation R
. Therefore, if you form
and do the inverse of the row operations which produced
in the reverse order, you would
By the same reasoning above, it follows that A
is a right inverse of B
also. It follows from Proposition 2.1.23
. Thus the procedure produces the
inverse whenever it works.
If it is possible to do row operations and end up with A
then the above
argument shows that A
has an inverse. Conversely, if A
has an inverse, can it be found by the
above procedure? In this case there exists a unique solution x
to the equation Ax
. In fact it
is just x
. Thus in terms of augmented matrices, you would expect to
That is, you would expect to be able to do row operations to A and end up with I.
The details will be explained fully when a more careful discussion is given which is based on
more fundamental considerations. For now, it suffices to observe that whenever the above
procedure works, it finds the inverse.
Example 2.1.26 Let A =
. Find A−1
Form the augmented matrix
Now do row operations until the n×n matrix on the left becomes the identity matrix. This yields
after some computations,
and so the inverse of A is the matrix on the right,
Checking the answer is easy. Just multiply the matrices and see if it works.
Always check your answer because if you are like some of us, you will usually have made a
Example 2.1.27 Let A =
. Find A−1
Set up the augmented matrix
times the first row and add to the second followed by
times the first row
added to the last. This yields
Then take 5 times the second row and add to −2 times the last row.
Next take the last row and add to
times the top row. This yields
times the second row and add to the top.
Finally divide the top row by −7, the second row by -10 and the bottom row by 14 which
Therefore, the inverse is
Example 2.1.28 Let A =
. Find A−1
Write the augmented matrix
and proceed to do row operations attempting to obtain
top row and add to the second. Then take
times the top row and add to the
times the second row to the bottom row.
At this point, you can see there will be no inverse because you have obtained a row of zeros in the
left half of the augmented matrix
Thus there will be no way to obtain I
on the left. In
other words, the three systems of equations you must solve to find the inverse have no
solution. In particular, there is no solution for the first column of A−1
because a sequence of row operations leads to the impossible equation, 0x + 0y + 0z = −1.