MA 265 LECTURE NOTES: FRIDAY, FEBRUARY 1
Finding Inverses of Matrices (cont’d)
Computing Inverses.
In the last lecture, we showed that an
n
×
n
matrix
A
is invertible if and only if
A
is
row equivalent to the
n
×
n
identity matrix
I
n
. In other words, say that there is a sequence of elementary row
operations which brings
A
into its reduced row echelon form
I
n
. These elementary row operations correspond
to elementary matrices
E
i
, so that
I
n
=
E A
in terms of
E
=
E
1
E
2
· · ·
E
k
.
Then
A

1
=
E
is the inverse of
A
.
In a more practical sense, we may keep track of these elementary row operations by considering the
n
×
2
n
augmented matrix:
bracketleftbig
A
I
n
bracketrightbig
=
a
11
a
12
· · ·
a
1
n
1
0
· · ·
0
a
21
a
22
· · ·
a
2
n
0
1
· · ·
0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
a
n
1
a
n
2
· · ·
a
nn
0
0
· · ·
1
.
We perform a series of elementary row operations to place this matrix in reduced row echelon form:
bracketleftbig
I
n
E
bracketrightbig
=
1
0
· · ·
0
e
11
e
12
· · ·
e
1
n
0
1
· · ·
0
e
21
e
22
· · ·
e
2
n
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0
0
· · ·
1
e
n
1
e
n
2
· · ·
e
nn
.
Example.
Consider the following 3
×
3 matrix:
A
=
1
1
1
0
2
3
5
5
1
We show how to compute
A

1
. We form the 3
×
6 augmented matrix:
bracketleftbig
A
I
3
bracketrightbig
=
1
1
1
1
0
0
0
2
3
0
1
0
5
5
1
0
0
1
We proceed to perform GaussJordan Reduction in order to compute the reduced row echelon form for this
matrix. First, we eliminate the entries in the first column:
(
−
5)
r
1
+
r
3
→
r
3
:
1
1
1
1
0
0
0
2
3
0
1
0
0
0
−
4
−
5
0
1
Now divide the second row by 2 in order to find the “leading one” for that row:
(1
/
2)
r
2
→
r
2
:
1
1
1
1
0
0
0
1
3
2
0
1
2
0
0
0
−
4
−
5
0
1
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 Spring '10
 ...
 Linear Algebra, Invertible matrix, Sn, elementary row operations, A. Permutations

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