If the last row of Arref is
all zeros, then matrix A is not
full rank;
and matrix A does not have an inverse.
If A is full rank, then the inverse of matrix
A is equal to the product of the elementary operators
that produced Arref , as shown below.
A-1 =
ErEr-1 . . .
E2E1
where
A-1 = inverse of matrix A r = Number of elementary row operations required to
transform A to
Arref Ei = ith elementary row operator
used to transform A to Arref
Note that the order in which elementary row operators are multiplied
is important, because
EiEj
is not necessarily equal to
EjEi.
An Example of Finding the Inverse
Let's use the above method to find the inverse of matrix
A, shown below.
A =
1
2
2
2
2
2
2
2
1
The first step is to transform matrix A into its
reduced row echelon form, Arref,
using a series of
elementary row operatorsEi. We show the transformation
steps below for each elementary row operator.
Multiply row 1 of A by -2 and add the
result to row 2 of A. This can be accomplished by pre-multiplying A by
the elementary row operator E1, which produces A1.
E1 =
1
0
0
-2
1
0
0
0
1
A1 = E1A =
1
2
2
0
-2
-2
2
2
1
Multiply row 1 of A1 by -2 and add the
result to row 3 of A1.
E2 =
1
0
0
0
1
0
-2
0
1
A2 = E2A1 =
1
2
2
0
-2
-2
0
-2
-3
Multiply row 3 of A2 by -1 and add
row 2 of A2 to
row 3 of A2.
E3 =
1
0
0
0
1
0
0
1
-1
A3 = E3A2 =
1
2
2
0
-2
-2
0
0
1
Add row 2 of A3 to
row 1 of A3.
E4 =
1
1
0
0
1
0
0
0
1
A4 = E4A3 =
1
0
0
0
-2
-2
0
0
1
Multiply row 2 of A4 by -0.5.
E5 =
1
0
0
0
-0.5
0
0
0
1
A5 = E5A4 =
1
0
0
0
1
1
0
0
1
Multiply row 3 of A5 by -1
and add the result to row 2 of A5.
The last matrix in Step 6 of the above table is
Arref, the reduced row echelon form for
matrix A. Since Arref
is equal to the identity matrix, we know that A is
full rank.
And because A is full rank, we know that A
has an inverse.
If A were less than full rank,
Arref would have all zeros in the last row; and
A would not have an
inverse.
We find the inverse of matrix A
by computing the product of the elementary operators
that produced Arref , as shown below.
A-1 =
E6E5E4E3E2E1
A-1 =
-1
1
0
1
-1.5
1
0
1
-1
In this example, we used a 3 x 3 matrix to show how to find a matrix inverse.
The same process will work on a square matrix of any size.
Multiply row 1 of A by -2 and add the
result to row 2 of A.
E1 =
1
0
-2
1
A1 = E1A =
1
0
0
2
Multiply row 2 of A1 by 0.5..
E2 =
1
0
0
0.5
Arref = E2A1 =
1
0
0
1
The last transformed matrix in the above table is
Arref,
the reduced row echelon form for
matrix A. Since the
reduced row echelon form is equal to the
identity matrix,
we know that A is
full rank.
And because A is full rank, we know that A
has an inverse.
We find the inverse by computing the product of the elementary operators
that produced Arref , as shown below.