In the last example, how would you find A^{−1}? You wish to find a matrix
( x z )
y w
such
that
( ) ( ) ( )
1 1 x z = 1 0 .
1 2 y w 0 1
This requires the solution of the systems of equations,
x + y = 1,x +2y = 0
and
z + w = 0,z + 2w = 1.
Writing the augmented matrix for these two systems gives
( )
1 1 | 1
1 2 | 0 (5.8)
(5.8)
for the first system and
( )
1 1 | 0
1 2 | 1 (5.9)
(5.9)
for the second. Lets solve the first system. Take
(− 1)
times the first row and add to the second to
get
( )
1 1 | 1
0 1 | − 1
Now take
(− 1)
times the second row and add to the first to get
( )
1 0 | 2
0 1 | − 1 .
Putting in the variables, this says x = 2 and y = −1.
Now solve the second system, 5.9 to find z and w. Take
(− 1)
times the first row and add to the second
to get
( )
1 1 | 0
.
0 1 | 1
Now take
(− 1)
times the second row and add to the first to get
( )
1 0 | − 1
0 1 | 1 .
Putting in the variables, this says z = −1 and w = 1. Therefore, the inverse is
( 2 − 1 )
.
− 1 1
Didn’t the above seem rather repetitive? Note that exactly the same row operations were used in both
systems. In each case, the end result was something of the form
(I|v)
where I is the identity and v gave a
column of the inverse. In the above,
( )
x
y
, the first column of the inverse was obtained first and then the
second column
( )
z
w
.
To simplify this procedure, you could have written
( )
1 1 | 1 0
1 2 | 0 1
and row reduced till you obtained
( )
1 0 | 2 − 1
0 1 | − 1 1
and read off the inverse as the 2 × 2 matrix on the right side.
This is the reason for the following simple procedure for finding the inverse of a matrix. This procedure
is called the Gauss-Jordan procedure.
Procedure 5.4.6Suppose A is an n × n matrix. To find A^{−1}if it exists, form the augmented n × 2nmatrix
(A|I)
and then, if possible do row operations until you obtain an n × 2n matrix of the form
(I|B). (5.10)
(5.10)
When this has been done, B = A^{−1}. If it is impossible to row reduce to a matrix of the form
(I|B)
, then Ahas no inverse.
Actually, all this shows is how to find a right inverse if it exists. What has been shown from the above
discussion is that AB = I. Later, I will show that this right inverse is the inverse. See Corollary 22.1.15 or
Theorem ?? presented later. However, it is not hard to see that this should be the case as
follows.
The row operations are all reversible. If the row operation involves switching two rows, the reverse row
operation involves switching them again to get back to where you started. If the row operation involves
multiplying a row by a≠0, then you would get back to where you began by multiplying the row by 1∕a. The
third row operation involving addition of c times row i to row j can be reversed by adding −c times row i
to row j.
In the above procedure, a sequence of row operations applied to I yields B while the same sequence of
operations applied to A yields I. Therefore, the sequence of reverse row operations in the opposite order
applied to B will yield I and applied to I will yield A. That is, there are row operations which
provide
(B|I) → (I|A)
and as just explained, A must be a right inverse for B. Therefore, BA = I. Hence B is both a right and a
left inverse for A because AB = BA = I.
If it is impossible to row reduce
(A|I)
to get
(I|B )
, then in particular, it is impossible to row reduce A
to I and consequently impossible to do a sequence of row operations to I and get A. Later it will be made
clear that the only way this can happen is that it is possible to row reduce A to a matrix of the form
( C )
0
where 0 is a row of zeros. Then there will be no solution to the system of equations represented by
the augmented matrix
( )
C | 0
0 1
Using the reverse row operations in the opposite order on both matrices in the above, it follows that there
must exist a such that there is no solution to the system of equations represented by
(A |a)
. Hence A fails
to have an inverse, because if it did, then there would be a solution x to the equation Ax = a given by
A^{−1}a.
Always check your answer because if you are like some of us, you will usually have made a
mistake.
Example 5.4.10In this example, it is shown how to use the inverse of a matrix to find the solution to asystem of equations. Consider the following system of equations. Use the inverse of a suitable matrix to givethe solutions to this system.
( )
| x+ z = 1 |
( x − y+ z = 3 ) .
x + y− z = 2
The system of equations can be written in terms of matrices as
More simply, this is of the form Ax = b. Suppose you find the inverse of the matrix A^{−1}. Then you could
multiply both sides of this equation by A^{−1} to obtain
( )
x = A−1A x = A −1(Ax ) = A−1b.
This gives the solution as x = A^{−1}b. Note that once you have found the inverse, you can easily get the
solution for different right hand sides without any effort. It is always just A^{−1}b. In the given example, the
inverse of the matrix is
( )
0 12 12
|( 1 − 1 0 |)
1 1
1 −2 − 2
This was shown in Example 5.4.9. Therefore, from what was just explained, the solution to the given
system is