Recall that the inverse of an n × n matrix A is a matrix B such that
AB = BA = I
where I is the identity matrix. It was shown that an elementary matrix is invertible
and that its inverse is also an elementary matrix. Also the product of invertible
matrices is invertible and its inverse is the product of the inverses in the reverse order.
In this section, we consider the problem of finding an inverse for a given n × n
matrix.
Example A.8.1Let A =
( 1 1 )
1 2
. Show that
( 2 − 1 )
− 1 1
is the inverse of A.
To check this, multiply
( )( ) ( )
1 1 2 − 1 = 1 0 ,
1 2 − 1 1 0 1
and
( )( ) ( )
2 − 1 1 1 = 1 0 ,
− 1 1 1 2 0 1
showing that this matrix is indeed the inverse of A.
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 (1.12)
1 2 | 0
(1.12)
for the first system and
( 1 1 | 0 )
1 2 | 1 (1.13)
(1.13)
for the second. Let’s 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, (1.13) 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? 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
Then you could have read off the inverse as the 2 × 2 matrix on the right side. You should
be able to see that it is valid by adapting the argument used in the simple case
above.
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 A.8.2Suppose A is an n×n matrix. To find A^{−1}if it exists, form theaugmented n × 2n matrix
(A |I)
and then if possible, do row operations until you obtain an n × 2n matrix of theform
(I|B). (1.14)
(1.14)
When this has been done, B = A^{−1}. If it is impossible to row reduce to a matrix of the form
(I|B)
, then A has no inverse.
The procedure just described along with the preceding explanation shows that this
procedure actually yields a right inverse. This is a matrix B such that AB = I. We will
show in Theorem A.8.4 that this right inverse is really the inverse. This is a stronger result
than that of Lemma A.6.8 about the uniqueness of the inverse. For now, here is an
example.
What you have really found in the above algorithm is a right inverse. Is this right
inverse matrix, which we have called the inverse, really the inverse, the matrix which when
multiplied on both sides gives the identity?
Theorem A.8.4Suppose A,B are n×n matrices and AB = I. Then it followsthat BA = I also, and so B = A^{−1}. For n × n matrices, the left inverse, right inverseand inverse are all the same thing.
Proof. If AB = I for A,B n × n matrices, is BA = I? If AB = I, there exists a unique
solution x to the equation
Bx = y
for any choice of y. In fact,
x = A (Bx ) = Ay.
This means the row reduced echelon form of B must be I. Thus every column is a pivot
column. Otherwise, there exists a free variable and the solution, if it exists, would not be
unique, contrary to what was just shown must happen if AB = I. It follows that a right
inverse B^{−1} for B exists. The above procedure yields
( ) ( )
B I → I B−1 .
Now multiply both sides of the equation AB = I on the right by B^{−1}. Then
( )
A = A BB −1 = (AB )B−1 = B− 1.
Thus A is the right inverse of B, and so BA = I. This shows that if AB = I, then BA = I
also. Exchanging roles of A and B, we see that if BA = I, then AB = I. This proves the
theorem. ■
This has shown that in the context of n × n matrices, right inverses, left inverses and
inverses are all the same and this matrix is called A^{−1}.
The following corollary is also of interest.
Corollary A.8.5An n × n matrix A has an inverse if and only if the row reducedechelon form of A is I.
Proof. First suppose the row reduced echelon form of A is I. Then Procedure A.8.2 yields
a right inverse for A. By Theorem A.8.4 this is the inverse. Next suppose A has an inverse.
Then there exists a unique solution x to the equation
Ax = y
given by x = A^{−1}y. It follows that in the augmented matrix
(A|0)
there are no free variables,
and so every column to the left of the zero column is a pivot column. Therefore, the row
reduced echelon form of A is I. ■