MA 265 LECTURE NOTES: WEDNESDAY, JANUARY 23
Special Types of Matrices
Linear Systems and Inverses.
We return to the question of solving a system of linear systems:
a
11
x
1
+
a
12
x
2
+
· · ·
+
a
1
n
x
n
=
b
1
a
21
x
1
+
a
22
x
2
+
· · ·
+
a
2
n
x
n
=
b
2
.
.
.
.
.
.
.
.
.
.
.
.
a
m
1
x
1
+
a
m
2
x
2
+
· · ·
+
a
mn
x
n
=
b
m
Recall that we can express this system as a product of matrices:
A
=
a
11
a
12
...
a
1
n
a
21
a
22
...
a
2
n
.
.
.
.
.
.
.
.
.
.
.
.
a
m
1
a
m
2
...
a
mn
,
x
=
x
1
x
2
.
.
.
x
n
and
b
=
b
1
b
2
.
.
.
b
n
=
⇒
A
x
=
b
.
For simplicity, assume that
m
=
n
. We will show that if
A
is a nonsingular matrix, then the system has a
unique solution.
Recall that we say that
A
is
nonsingular
(or
invertible
) if there exists an
n
×
n
matrix
A

1
such that
AA

1
=
A

1
A
=
I
n
is the
n
×
n
identity matrix. First we show that the system has
at least
one solution
x
1
. Indeed, denote
x
1
=
A

1
b
. Then we have
A
x
1
=
A
(
A

1
b
)
=
(
AA

1
)
b
=
I
n
b
=
b
so that
x
1
is a solution.
Now we show that the system has
at most
one solution.
Indeed, say that
x
2
is
another solution. Then we have
x
2
=
I
n
x
2
=
(
A

1
A
)
x
2
=
A

1
(
A
x
2
) =
A

1
b
=
x
1
.
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 Spring '10
 ...
 Invertible matrix, Row echelon form, Inverses, Matrices Linear Systems

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