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 (18.25)
1 2 | 0
(18.25)
for the first system and
( )
1 1 | 0 (18.26)
1 2 | 1
(18.26)
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, 18.26 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 18.6.6Suppose A is an n×n matrix. To find A^{−1}if it exists, formthe augmented n × 2n matrix
(A|I)
and then, if possible do row operations until you obtain an n × 2n matrix of theform
(I|B). (18.27)
(18.27)
When this has been done, B = A^{−1}. If it is impossible to row reduce to a matrix of theform
(I|B )
, then A has 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. In fact this is the inverse. 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 18.6.10In this example, it is shown how to use the inverse of a matrixto find the solution to a system of equations. Consider the following systemof equations. Use the inverse of a suitable matrix to give the solutions to thissystem.
( 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 1 1 )
( 1 −2 1 20 )
1 − 1 − 1
2 2
This was shown in Example 18.6.9. Therefore, from what was just explained, the solution
to the given system is