Math 334 Lecture #33
§
7.7: Fundamental Matrices
Example.
Recall that a general solution of
x
=
Ax
where
A
=
1
1
4
1
is
x
(
t
) =
c
1
x
(1)
+
c
2
x
(2)
=
c
1
e
3
t
2
e
3
t
+
c
2
e

t

2
e

t
=
e
3
t
e

t
2
e
3
t

2
e

t
c
1
c
2
.
The form of this general solution is that of an invertible (why?) 2
×
2 matrix function
times a vector of arbitrary constants,
x
(
t
) = Ψ(
t
)
c,
where the columns of the 2
×
2 the matrix function,
Ψ(
t
) =
e
3
t
e

t
2
e
3
t

2
e

t
,
are the linearly independent solutions
x
(1)
and
x
(2)
of the system.
The function Ψ(
t
) is called a
fundamental matrix
for the system; it satisfies the first
order differential matrix equation
Ψ (
t
) =
A
Ψ(
t
)
where
A
=
1
1
4
1
.
This is true because each column of Ψ(
t
) is a solution of the system. This can be verified:
Ψ (
t
) =
3
e
3
t

e

t
6
e
3
t
2
e

t
and
1
1
4
1
Ψ(
t
) =
1
1
4
1
e
3
t
e

t
2
e
3
t

2
e

t
=
3
e
3
t

e

t
6
e
3
t
2
e

t
.
The Form of General Solutions.
If
x
(1)
(
t
)
, . . . , x
(
n
)
(
t
) are
n
linearly independent
solutions of
x
=
P
(
t
)
x
where
P
(
t
) is an
n
×
n
matrix function, then a general solution
of the system is
x
(
t
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 Spring '11
 Smith
 Linear Algebra, Derivative, Matrices, Vector Space, Matrix exponential

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