Math 211 Fall 2007: Solutions: HW #13
Instructor: S. Cautis
1. section 9.5, #2
We have
det(A − λI)


−1 − λ
6
2
= det  0
−1 − λ
0 
−1
11
2−λ


0
−1 − λ
0
6
2 
= − det −1 − λ
−1
11
2−λ
= (−1 − λ)((−1 − λ)(2 − λ) + 2)
= −(1 + λ)(λ)(λ − 1)
where to get the second line we interchanged the first two rows (and hence
multiplied the determinant by −1 as required).
So the characteristic polynomial is p(λ) = −λ(λ − 1)(λ + 1). The roots of
the polynomial are 0, 1, −1 which correspond to the eigenvalues. So the
eigenvalues are the values where the graph of the characteristic polynomial
intersects the horizontal axis.
2. section 9.5, #4
The characteristic polynomial is p(λ) = (1 − λ)((−2 − λ)(2 − λ) + 5) =
(1 − λ)(λ2 + 1). So the eigenvalues are 1, i and −i.
For the first eigenvalue we get eigenvector (1, 0, 0)t . For the eigenvalue i
we get eigenvector (0, 1, 2 + i)t so for the eigenvalue −i (its conjugate) we
get eigenvector (0, 1, 2 − i)t . To see these vectors are linearly independent
we compute the determinant


1
0
0
1
1  = (2 − i) − (2 + i) = 2i
det 0
0 2+i 2−i
which is nonzero. This means that the vectors are linearly independent.
3. section 9.5, #10
We have
det(A − λI) = (−3 − λ)(1 − λ)(−1 − λ) + 6 · 0 − 2 · 0 = −(λ + 3)(λ − 1)(λ + 1)
so that the eigenvalues are −3, 1, −1. The corresponding eigenvectors are
(1, 0, 0)t , (1, 0, −1)t and (1, −1, 1)t .
Hence the general solution is
C1 e−3t (1, 0, 0)t + C2 et (1, 0, −1)t + C3 e−t (1, −1, 1)t.
4. section 9.5, #22 We have
det(A − λI) =
(2 − λ)((2 − λ)(−5 − λ) + 12) − 4((−5 − λ) + 9) + 4(−4 + 3(2 − λ))
=
=
(2 − λ)(λ2 + 3λ + 2) + 4λ − 16 − 12λ + 8
−λ3 − λ2 − 4λ − 4
=
−(λ + 1)(λ2 + 4)
So the eigenvalues are λ1 = −1 λ2 = 2i and λ3 = −2i.
An eigenvector for λ1 is v1 = (0, 1, −1)t . An eigenvector for λ2 is v2 =
(−2 − 4i, 1 − 3i, 2 + 4i)t while an eigenvector for λ3 is the conjugate v3 =
(−2 + 4i, 1 + 3i, 2 − 4i)t . So the general solution is
C1 e−t v1 + C2 e2i v2 + C3 e−2i v3
5. section 9.8, #2
(a) The substitution part is easy. To see that x1 and x2 are linearly
independent we compute the determinant
sin 2t
cos 2t
det
= −2 sin2 (2t) − 2 cos2 (2t) = −2
2 cos 2t −2 sin 2t
Since this is nonzero x1 and x2 are indeed linearly independent.
(b) The first component of C1 x1 + C2 x2 is C1 sin(2t) + C2 cos(2t) which
is easily checked to satisfy y ′′ + 4y = 0.
6. section 9.8, #8
The Wronskian of e−10t and et is
−10t
e
W (t) =
−10e−10t
Hence
W (0) =
et
et
1
1
−10 1
which has determinant 11 which is nonzero. Thus the solutions are linearly
independent.
7. section 9.8, #24
The characteristic polynomial is λ3 +6λ2 +12λ+8 = (λ+2)(λ2 +4λ+4) =
(λ + 2)3 which has a triple root of λ = −2. So the general solution is
C1 e−2t + C2 te−2t + C3 t2 e−2t
8. section 9.8, #40
The characteristic polynomial is λ3 − 2λ + 4 = (λ + 2)(λ2 − 2λ + 2) which
has roots −2 and 1 ± i. So a basis for the set of solutions is e−2t , and
the real and imaginary parts of e(1+i)t = et (cos(t) + i sin(t)). Hence the
general solution is
y = C1 e−2t + C2 et cos(t) + C3 et sin(t)
The initial conditions y(0) = 1 implies C1 + C2 = 1.
Now y ′ = −2C1 e−2t + C2 et cos(t) − C2 et sin(t) + C3 et sin(t) + C3 et cos(t)
so the initial conditions y ′ (0) = −1 implies −2C1 + C2 + C3 = −1.
Finally y ′′ = 4C1 e−2t +C2 et cos(t)−C2 et sin(t)−C2 et sin(t)−C2 et cos(t)+
C3 et sin(t)+C3 et cos(t)+C3 et cos(t)−C3 et sin(t) so the condition y ′′ (0) = 0
means 4C1 + C2 − C2 + C3 + C3 = 0 or 4C1 + 2C3 = 0 so C3 = −2C1 .
Substituting back gives −2C1 + C2 − 2C1 = −1 so −4C1 + C2 = −1. Thus
5C1 = 2 and C1 = 2/5 and C2 = 3/5 and C3 = −4/5.