Matrix Inverses

A Version of Inversion
(we hope not aversion to inversion!)

This centre was established to provide help to visitors who are not familiar with the ideas of the inverse of a matrix and/or with a method of finding an inverse using row operations.

Our aim is not to compete with the numerous books that cover this basic material. If you are really not familiar with the subject we suggest that you consult a book on this subject for more details.

Basic idea of inverses

Consider the two matrices.

A =
1 5 2
1 1 7
0 -3 4
B =
-25 26 -33
4 -4 5
3 -3 4

On the surface there seems to be no particular relationship between them. However if we compute AB we find that

AB =
1 0 0
0 1 0
0 0 1

that is we get the 3x3 identity matrix, I3.

If you are not familiar with the famous matrix identity I, visit the tutOR Module Matrix Basics. If you don't know how to multiply matrices see the tutOR Module Matrix_Product.

We now make an analogy with familiar ideas of multiplication of real numbers, where we know that, if ax = 1, then x = a-1. We say a-1 is the multiplicative inverse of a. In the example above, because AB = I3, using this analogy, we say
B = A-1 and we call A-1 the inverse of A.

Definition of the Inverse of a Matrix

Let A and B be nxn matrices. If there is a B such that

AB = BA = In

then B is called the inverse of A, written A-1. A is then said to be invertible. We usually just use I rather than In if the size is clear from the context.

So we then have

A A-1 = A-1A = I.

NOTES

  • Inverses are only defined for square matrices.
  • An identity crisis:
    Many square matrices do not have an inverse. (Compare with real numbers: 0 has no multiplicative inverse.)

    • You

    can't

    invert

    me!

  • If a square matrix A has an inverse, then this inverse is unique. That is, there is only one matrix B such that
    AB = BA = I.
  • Although with real numbers we use the notation a-1 = 1/a, we do not use this notation with matrix inverses. We do not have a concept of dividing matrices.

    • Finding Inverses

    There are several techniques for finding inverses. The one we present here we call Exchanging Identities. This method tells us whether or not an inve rse exists as well as finding it when it does exist. For a discussion of other methods and also to see why the method we call Exchanging Identities works, see a basic Linear Algebra textbook.

    The Method of Exchanging Identities for Finding Inverses

    1. Given an nxn matrix A, we create the augmented matrix (A | I ) by appending the nxn identity matrix to the right of A, with a separating line shown here in green.

    For example if

    A =
    1 5 2
    1 1 7
    0 -3 4

    we consider

    (A | I) =
    1 5 2 | 1 0 0
    1 1 7 | 0 1 0
    0 -3 4 | 0 0 1

    2. We then use row operations to try to reduce the matrix so that we get the identity on the left where we originally had A . If you don't know about row operations, go to the tutOR Module Row Operations.

    3. If we can do this, the matrix is invertible AND the matrix appearing in the right where we originally had I is now A-1. That is (A | I ) reduces to (I |A-1 ). Isn't that beautiful!!

    4. If we get a row of zeros in the left where we originally had A, then A is not invertible. There are other simple ways of showing whether a matrix has an inverse, see for example tutOR Module: Determi nants.

    EXAMPLE: Find the inverse of the matrix A given below.

    A =
    1 5 2
    1 1 7
    0 -3 4

    We consider

    (A | I) =
    1 5 2 | 1 0 0
    1 1 7 | 0 1 0
    0 -3 4 | 0 0 1

    There are many ways of proceeding. We describe here two distinct sequences of row operations, which we will see lead us to the same inverse, as of course they should, since, as we stated earlier, the inverse is unique.

    Each of the versions we describe has its advantages.The second is better, in this case, if you are doing it by hand, as it avoids the use of fractions.

    Remember you can look at tutOR Module Row Operations. if you don't know what is going on here.

    FIRST VERSION

    Pivot on the one in the first row and first column. This gives the row equivalent matrix

    1 5 2 | 1 0 0
    0 -4 5 | -1 1 0
    0 -3 4 | 0 0 1

    Now pivoting on the -4 in the second row and second column, we get

    1 0 33/4 | -1/4 5/4 0
    0 1 -5/4 | 1/4 -1/4 0
    0 0 1/4 | 3/4 -3/4 1

    Finally pivoting on the 1/4 in the third row and third column gives

    1 0 0 | -25 26 -33
    0 1 0 | 4 -4 5
    0 0 1 | 3 -3 4

    Pivorrrratti wipes brow and rests. From the right side we have

    A-1 =
    -25 26 -33
    4 -4 5
    3 -3 4

    We could now check and make sure that A-1A = I.

    SECOND VERSION

    Pivot on the one in the first row and first column. This gives the row equivalent matrix

    1 5 2 | 1 0 0
    0 -4 5 | -1 1 0
    0 -3 4 | 0 0 1

    Now subtract row 3 from row 2. This gives the row equivalent matrix

    1 5 2 | 1 0 0
    0 -1 1 | -1 1 -1
    0 -3 4 | 0 0 1

    Pivot on the -1 in the second row and second column. We get

    1 0 7 | -4 5 -5
    0 1 -1 | 1 -1 1
    0 0 1 | 3 -3 4

    Finally pivot on the one in the third row and third column to give

    1 0 0 | -25 26 -33
    0 1 0 | 4 -4 5
    0 0 1 | 3 -3 4

    Thus from the right side, we have as before

    A-1 =
    -25 26 -33
    4 -4 5
    3 -3 4

    ANOTHER EXAMPLE: CAN YOU INVERT ME?

    Consider the matrix A and its corresponding augmented matrix

    A =
    1 2
    -3 -6
    (A | I) =
    1 2 | 1 0
    -3 -6 | 0 1

    Adding three times row 1 to row 2 we get

    1 2 | 1 0
    0 0 | 3 1

    Now because of the row of zeros in the left section, there is no way we can get the 2x2 identity matrix there. So this matrix does not have an inverse. A does not have a companion matrix to give an identity.

    TO SUMMARIZE

    Given a matrix A :

  • Consider the augmented matrix (A | I).
  • Use row operations to transform this to (I |B), if possible.
  • Then B = A-1.
  • If we obtain a row of zeros to the left of the vertical line, the inverse of A does not exist.