Math 220
Robert Creese
September 12, 2019
Robert Creese
September 12, 2019
1 / 32
1.4 The Matrix Equation:
A
x
=
b
Robert Creese
September 12, 2019
2 / 32
1.4 Product of a matrix and vector
A
x
Let
A
be a
m
×
n
matrix, with each column denoted by
a
1
,
a
2
,
· · ·
,
a
n
.
A
=
a
1
a
2
· · ·
a
n
If
x
is in
R
n
, then we can define the product of
A
times
x
written
A
x
, as
the linear combination of the columns of
A
using the corresponding entires
of
x
as weights.
A
x
=
a
1
a
2
· · ·
a
n
x
1
x
2
.
.
.
x
n
=
x
1
a
1
+
x
2
a
2
+
· · ·
+
x
n
a
n
This is only defined when
A
has the same number of columns as
x
has
entries (rows).
Robert Creese
September 12, 2019
3 / 32
1.4 Example: Multiplying a matrix and vector
To evaluate a matrix times a vector we can write the equivalent expression
using vectors and evaluate the expression as in Sec. 1.3.
4
2

1
3
2
5
1

2
= 1
4

1
2
+ (

2)
2
3
5
=
4

1
2
+

4

6

10
=
0

7

8
2
4
1
2

1
3
4

3
5
= 4
2
2
+ (

3)
4

1
+ 5
1
3
=
1
26
Robert Creese
September 12, 2019
4 / 32
1.4 Vector Operations as a matrix times a vector
Given
v
1
,
v
2
,
v
3
,
v
4
write 2
v
1
+ 4
v
2

9
v
3
+
v
4
as a matrix times a vector.
Just use the equation definition of a matrix times a vector
a
1
a
2
· · ·
a
n
x
1
x
2
.
.
.
x
n
=
x
1
a
1
+
x
2
a
2
+
· · ·
+
x
n
a
n
. This leads to
2
v
1
+ 4
v
2

9
v
3
+
v
4
=
v
1
v
2
v
3
v
4
2
4

9
1
Robert Creese
September 12, 2019
5 / 32
1.4 Matrix Equation
Given any linear system an equivalent vector equation can be found, for
example,
x
1

3
x
2
+ 5
x
3
= 1
2
x
1
+
x
2

4
x
3
=

3
is equivalent to
x
1
1
2
+
x
2

3
1
+
x
3
5

4
=
1

3
This is also a
matrix equation
, an equation written as
A
x
=
b
1

3
5
2
1

4
x
1
x
2
x
3
=
1

3
Robert Creese
September 12, 2019
6 / 32
1.4 Matrix Equation: Finding a solution set
If
A
is a
m
×
n
matrix with columns
a
1
, . . . ,
a
n
and
b
in
R
m
, the
matrix
equation
A
x
=
b
has the same solution set as the linear system whose augmented matrix is
a
1
. . .
a
n
b
The main method to solve any matrix equation
A
x
=
b
is to find the
reduced echelon form of the corresponding augmented matrix.
Robert Creese
September 12, 2019
7 / 32
1.4 Existence of solutions to
A
x
=
b
A
x
=
b
has a solution if and only if
b
is a linear combination of the
columns of
A
,
{
a
1
, . . . ,
a
n
}
Section 1.3 covered the problem is a specific
b
in Span
{
a
1
, . . . ,
a
n
}
? This
is equivalent problem to asking if
A
x
=
b
has at least one solution.
A new problem to consider:
Is
A
x
=
b
consistent for all possible
b
?
Robert Creese
September 12, 2019
8 / 32
1.4 Example: Consistency
Does
A
x
=
b
have a solution for all possible
b
where
A
=
2
1
4
6
?
Begin by writing the corresponding augmented matrix (
b
= (
b
1
,
b
2
)) and
finding echelon form
2
1
b
1
4
6
b
2
Replace row 2 by
r
2

2
r
1
2
1
b
1
0
4
b
2

2
b
1
The echelon form is consistent. Therefore
A
x
=
b
has a solution for all
possible
b
in
R
2
.
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