PROBLEM SET
Problems on Matrix Exponentials and Differential equations
Corrected Version!
Math 3351, Fall 2010
Oct. 28, 2010
ANSWERS
i
Problem 1.
Consider the system of differential equations
x01 (t) = 15x1 (t) + 16x2 (t)
(1)
x02 (t) = −12x1 (t) − 13x2 (t)
A. Express the system in the matrix form x0 = Ax.
Answer :
We can write the system as
0 0 x1 (t)
x1 (t)
15
16
x1 (t)
= 0
=
x2 (t)
x2 (t)
−12 −13 x2 (t)
This is in the form x0 = Ax with
15
16
A=
.
−12 −13
B. Compute the matrix etA .
Answer :
The matrix A appeared on the second set of practice problems, where we
found the characteristic polynomial and found that the eigenvalues of A are
−1 and 3. Since there are two distinct eigenvalues of the 2 × 2 matrix A, A
must be diagonalizable.
Let’s start with the eigenvalue λ = −1. We compute that
"
#
16
16
A − (−1)I = A + I =
.
−12 −12
Using a calculator, we find the RREF of A + I to be
"
#
1 1
R=
.
0 0
We need to find the nullspace of R. Call the variables x1 and x2 . From R,
we see that x1 is a leading variable and x2 is a free variable, say x2 = α.
The first row of R tells us that
x1 + x2 = 0 =⇒ x1 = −x2 = −α.
Thus, the nullspace of R is parametrized by
"
# "
#
"
#
x1
−α
−1
=
=α
x2
α
1
1
We conclude that the eigenspace E(−1) is one dimensional with basis vector
"
#
−1
v1 =
1
Next, we do the eigenvalue λ = 3. We compute
"
#
12
16
A − 3I =
−12 −16
The RREF of A − 3I is
"
1
4/3
0
0
R=
#
.
Again, x1 is a leading variable and x2 is free, say x2 = α. The first row of
R tells us that
4
4
4
x1 + x2 = 0 =⇒ x1 = − x2 = − α.
3
3
3
Thus, the nullspace of R is parametrized by
"
# " 4 #
"
#
x1
−4/3
−3α
=
=α
.
x2
1
α
Thus, we conclude that the eigenspace E(3) is one dimensional with basis
vector
"
#
−4/3
u=
.
1
We can get rid of the fraction here by noting that we can multiply every
vector in a basis by a nonzero number to get a new basis. Thus, instead of
using u as the basis of E(3), we can use the vector v2 = 3u. Thus, we have
"
#
−4
v2 =
3
as the basis vector for E(3). Thus, we have two eigenvectors v1 and v2 ,
which we know form a basis of three dimensional space.
To diagonalize A, we form the matrix P with columns v1 and v2 , so
"
#
−1 −4
P =
.
1
3
We form the diagonal matrix D, where the diagonal entry in each column
of D is the eigenvalue that goes with the corresponding entry of P . Thus,
"
#
−1 0
D=
.
0 3
2
You can use your calculator to check P −1 AP = D.
Next, we want to compute etA . The rule for computing the exponential of
a diagonal matrix gives us
" −t
#
e
0
tD
e =
0 e3 t
Since A = P DP −1 , we have P etD P −1 = etP DP
−1
= etA Thus,
etA = P etD P −1
"
# " −t
#"
#
−1 −4
e
0
3
4
=
1
3
0 e−3 t
−1 −1
"
#
−3 e−t + 4 e3 t −4 e−t + 4 e3 t
=
3 e−t − 3 e3 t
4 e−t − 3 e3 t
C. Find the solution of the system (1) for arbitrary initial conditions
x1 (0) = x01
x2 (0) = x02 .
Answer :
Let
x0 =
0
x1
x02
Then the solution of the system x0 = Ax with initial conditions x(0) = x0
is x(t) = etA x0 . So, in our case,
"
# "
#" 0 #
x1 (t)
−3 e−t + 4 e3 t −4 e−t + 4 e3 t
x1
=
x2 (t)
3 e−t − 3 e3 t
4 e−t − 3 e3 t
x02
"
#
−3 e−t + 4 e3 t x01 + −4 e−t + 4 e3 t x02
=
3 e−t − 3 e3 t x01 + 4 e−t − 3 e3 t x02
D. find the solution of the system (1) for the specific initial conditions
x1 (0) = 2
x2 (0) = −5.
Answer :
Substitute x01 = 2 and x02 = −5 in the last answer. After simplification, we
get
"
# "
#
x1 (t)
14 e−t − 12 e3 t
=
x2 (t)
−14 e−t + 9 e3 t
3
Problem 2. In each part, diagonalize the matrix A and compute etA . (These
matrices were diagonalized on a previous set of practice problems!) In the case of complex eigenvalues, present etA with entries that are
obviously real (i.e., not is appear).
A.

−16
36 −18

A=
 −6
14
3
−6


−6 
.
5
Answer :
We diagonalized this matrix in the previous problem set, where we found


−6 −1 2


0 1 
P =

 −2
1
1 0


−1 0 0



D=
 0 2 0 .
0 0 2
We then have

e−t

etD = 
 0
0
0
e2 t
0

0

0 

2t
e
and

6 e−t − 5 e2 t

−t
2t
etD = P etD P −1 = 
 2e − 2e
−e−t + e2 t
−12 e−t + 12 e2 t
−4 e−t + 5 e2 t
2 e−t − 2 e2 t
which I got by using a calculator.
B.

−2
−4

A=
 −3
−3
−5
4
5


5 

−5 8
6 e−t − 6 e2 t


2 e−t − 2 e2 t 
,
−t
2t
−e + 2 e
Answer :
We diagonalized this matrix on a previous problem set. The eigenvalues are
1, 1 + i and 1 − i, and we found


3 7−i 7+i



P =
 4 7−i 7+i 
5
10
10


1
0
0


0 
D=
.
 0 1+i
0
0
1−i
We can easily find

et

etD = 
 0
0
0
0

e(1+i)t
0
0
e(1−i)t

,

but to get the final answers to come out in real form, we need to write out
the complex exponentials in terms of their real and imaginary parts. Thus,
we should write

 t
e
0
0


t
t

0
e
cos
(t)
+
ie
sin
(t)
0
etD = 


0
0
et cos (t) − iet sin (t)
Now, we can compute (by calculator!)
etA == P etD P −1

−3 et + 4 et cos (t) − 3 et sin (t) 3 et − 3 et cos (t) − 4 et sin (t)

t
t
t
t
t
t
=
 −4 e + 4 e cos (t) − 3 e sin (t) 4 e − 3 e cos (t) − 4 e sin (t)
−5 et + 5 et cos (t) − 5 et sin (t) 5 et − 5 et cos (t) − 5 et sin (t)

5 et sin (t)

.

et cos (t) + 7 et sin (t)
Problem 3. This problem deals with finding etA and solving the system x0 =
Ax in an example where A is not diagonalizable. Let A be given
by


−2 7 −10


.
−2
6
−4
A=


−1 3
2
5
5 et sin (t)
The characteristic polynomial of A is −(λ − 2)3 , so there is only one eigenvalue,
namely λ = 2. The eigenspace E(2) is only one dimensional and has a basis
given by the vector


−6



v1 = 
 −2  .
1
Thus, A is certainly not diagonalizable.
To find etA and solve the system x0 = Ax, proceed as follows.
A. Solve the system (A − 2I)x = v1 . You’ll get a one parameter family of
solutions. Pick a specific solution and call it v2 (so v2 is a constant vector).
Thus, (A − 2I)v2 = v1 .
B. Similarly, find a specific vector v3 so that (A − 2I)v3 = v2 .
C. Let P be the matrix whose columns are v1 , v2 and v3 . Calculate J = P −1 AP .
D. Solve the system of differential equations z 0 = Jz by hand (work from the
bottom up) with initial conditions z 0 , and thus find etJ .
E. Calculate etA = P etJ P −1 . This tells you how to solve the system x0 = Ax,
of course.
Answer :
I get


etA = 

e2 t − 4 te2 t + 6 t2 e2 t
7 te2 t − 15 t2 e2 t
−2 te2 t + 2 t2 e2 t
e2 t + 4 te2 t − 5 t2 e2 t
−te2 t − t2 e2 t
3 te2 t + 25 t2 e2 t
6
−10 te2 t + 6 t2 e2 t


−4 te2 t + 2 t2 e2 t 

2 2t
2t
−t e + e