**Unformatted text preview: **yick (tay236) – HW 10 – gilbert – (53415)
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001 2
1 −1
4 10.0 points An n × n matrix can be diagonalizable, but
not invertible.
True or False? is diagonalizable, i.e., A = P DP −1 with P
invertible and D diagonal, which of the following is a choice for D?
−3 0
1. D =
0 2
3 0
2. D =
0 3
3 0
3. D =
0 2
−3 0
4. D =
0 −3
5. A is not diagonalizable correct
Explanation:
Since
det [A − λI] = A is not diagonalizable . 002 If the matrix
A = Consequently, 10.0 points 1 1. FALSE
2. TRUE correct
Explanation:
Consider the 3 × 3 triangular matrix 5 −8 1
A = 0 0
7 0 0 −2 Because A is triangular, its eigenvalues are the
entries along the diagonal, i.e., λ = 5, 0, −2.
Since these are distinct, A is diagonalizable.
On the other hand, one of its eigenvalues
is zero, so A is not invertible (or note that
det[A] = 0 because
det[A] = 5(0)(−2) = 0 2−λ
1 −1
4−λ is the product of the diagonal values of A, so A
is not invertible). Therefore, an n × n matrix
A can be diagonalizable, but not invertible. = 1 + (2 − λ)(4 − λ) = 9 − 6λ + λ2 , Consequently, the statement is the eigenvalues of A are the solutions of
9 − 6λ + λ2 = (3 − λ)2 = 0 , TRUE . i.e., λ = 3, 3.
On the other hand, when λ = 3,
1
−1 −1
=
rref(A − λI) = rref
0
1
1 003
1
0 , so x2 is the only free variable. Thus the
eigenspace Nul(A − 3I) has dimension 1. But
then, when λ = 3,
geo multA (λ) < alg multA (λ) . 10.0 points If the matrix
A = 1
1 4
−2 is diagonalizable, i.e., A = P DP −1 with P
invertible and D diagonal, which of the following is a choice for P ? yick (tay236) – HW 10 – gilbert – (53415)
1. P = 2. P = 2
−1
−1
2 4
1 4
1 which can be written as
x1 + 4x2 = 2x1 , 4. P = −1
1 5. P = −1
1
4
correct
1
1
4 Consequently, A = P DP −1 with
P = Explanation:
Since 004 det [A − λI] = 1−λ
1 4
−2 − λ = −4 − (1 − λ)(2 + λ) = λ2 + λ − 6 ,
the eigenvalues of A are the solutions of
λ2 + λ − 6 = (λ + 3)(λ − 2) = 0 ,
i.e., λ = −3, 2. Thus A is diagonalizable
because the eigenvalues of A are distinct, and
A = P DP −1 with which can be written as −1
1 4
1 . 10.0 points 1. FALSE correct
2. TRUE
Explanation:
If an n×n matrix has n distinct eigenvalues,
it is diagonalizable, but the converse does not
necessarily have to be true.
For example, when 1 A = −3
3 where v1 and v2 are eigenvectors corresponding to λ1 and λ2 respectively. To determine
v1 and v2 we solve the equation Ax = λx.
Ax = λx, λ = −3:
x
x1 + 4x2
x1
1 4
= −3 1 ,
=
x2
x1 − 2x2
x2
1 −2 If an n × n matrix A is diagonalizable, then
A has n distinct eigenvalues.
True or False? P = [v1 v2 ] x1 + 4x2 = −3x1 , x1 − 2x2 = 2x2 , i.e., x1 = 4x2 . So one choice of v2 is
4
.
v2 =
1 3. A is not diagonalizable
2 3
−5
3 then 3
−3 ,
1 det[A − λI] = −(λ − 1)(λ + 2)2 = 0.
Thus the eigenvalues of A are λ = 1, −2, −2. x1 − 2x2 = −3x2 , i.e., x1 = −x2 . So one choice of v1 is
−1
.
v1 =
1
Ax = λx, λ = 2:
x
x1 + 4x2
x1
1 4
= 2 1 ,
=
x2
x1 − 2x2
x2
1 −2 Now so 1
rref(A − I) = 0
0 0 −1
1 1 ,
0 0 1 Nul(A − I) = s −1 : s in R . 1 yick (tay236) – HW 10 – gilbert – (53415)
On the other hand, so 1 1
rref(A + 2I) = 0 0
0 0 1
0,
0 Nul(A + 2I) −1 −1
= s 1 + t 0 : s, t in R . 0
1 But then A is diagonalizable because it has 3
linearly independent eigenvectors −1
−1
1 −1 , 1 , 0 .
0
1
1 Since the eigenvalue λ = −2 is repeated, however, A does not have distinct eigenvalues.
Consequently, the statement is
FALSE . 005 10.0 points The eigenvalues of the matrix 2 0 −2
A = 1 3 2 0 0 3 are λ = 2, 3, 3. If A is diagonalizable, i.e., A = P DP −1
with P invertible and D diagonal, which of
the following is a choice for P ? 1 0 −2
1. P = 1 1 0 0 0 1
2. A is not diagonalizable −1
3. P = 1
0 0
1
0 2
0
1 3 1 0 2
4. P = 1 1 0 0 0 1 −1 0 −2
5. P = 1 1 0 correct
0 0 1 Explanation:
We first determine the eigenspaces corresponding to λ = 2, 3:
λ = 2 : since 0 0 −2
rref(A − 2I) = rref 1 1 2 0 0 1 1 1 0
= 0 0 1,
0 0 0 there is only free variable and −1 Nul(A − 2I) = s 1
: s in R . 0 Thus Nul(A − 2I) has dimension 1, and −1
v1 = 1 0 is a basis for Nul(A − 2I).
λ = 3 : since −1 0
rref(A − 3I) = rref 1 0
0 0 1 0 2
= 0 0 0,
0 0 0 −2
2 0 there are two free variables and −2 0 Nul(A−3I) = s 1 + t 0 : s, t in R . 0
1
Thus Nul(A − 3I) has dimension 2, and −2
0
v2 = 1 , v3 = 0 ,
0
1 yick (tay236) – HW 10 – gilbert – (53415)
form a basis for Nul(A − 3I).
Consequently, A is diagonalizable because
v1 , v2 , v3 are linearly independent, and A =
P DP −1 with −1 0 −2
P = [v1 v2 v2 ] = 1 1 0 .
0 0 1
006 4 On the other hand, one of its eigenvalues
is zero, so A is not invertible (or note that
det[A] = 0 because det[A] is the product
5(0)(−2) = 0 of the diagonal values of A,
hence not invertible). Therefore, an n × n
matrix A can be diagonalizable, but not invertible.
Consequently, the statement is 10.0 points
FALSE . Every n × n matrix A having linearly independent eigenvectors v1 , v2 , . . . , vn can be
diagonalized. 008 By diagonalizing the matrix True or False?
1. TRUE correct A = 2. FALSE
Explanation:
An n × n matrix A can be diagonalized,
i.e. written as A = P DP −1 for some invertible matrix P and diagonal matrix D, if and
only if A has linearly independent eigenvectors
v1 , v2 , . . . , vn .
Consequently, the statement is
TRUE .
007 10.0 points 10.0 points If A is a diagonalizable n × n matrix, then
A is invertible.
True or False?
1. TRUE
2. FALSE correct
Explanation:
Consider the 3 × 3 triangular matrix 5 −8 1
A = 0 0
7 0 0 −2 Because A is triangular, its eigenvalues are the
entries along the diagonal, i.e., λ = 5, 0, −2.
Since these are distinct, A is diagonalizable. 3
2
−4
,
−3 compute f (A) for the polynomial
f (x) = 3x3 − 3x2 + 3x − 2 . 1. f (A) =
2. f (A) =
3. f (A) =
4. f (A) =
13 −24
correct
12 −23
−13 24
−12 23
13 −24
−12 23
−13 24
12 −23
Explanation:
If A can be diagonalized by
A = P DP −1 = P d1
0
0
P −1 ,
d2 then
f (A) = P f (D)P −1
f (d1 )
0
P −1 .
= P
0
f (d2 ) yick (tay236) – HW 10 – gilbert – (53415)
Now A can be diagonalized if we can find
an eigenbasis of R2 of eigenvectors v1 , v2 of A
corresponding to eigenvalues λ1 , λ2 , for then:
λ1 0
P −1 , P = [v1 v2 ] .
A = P
0 λ2 1. FALSE
2. TRUE correct
Explanation:
The eigenspaces But 3 − λ
det[A − λI] = 2 Nul(A − λ1 I), −4 −3 − λ Nul(A − λ1 I), Nul(A − λ2 I), Nul(A − λ3 I), i.e., λ1 = 1 and λ2 = −1. Corresponding
eigenvectors are
1
2
,
,
v2 =
v1 =
1
1 must be at least four.
On the other hand, since A is 4 × 4, the sum
of the dimensions of
Nul(A − λ1 I), Nul(A − λ2 I), Nul(A − λ3 I), so
P = 2 1
1 1 2
1 , P −1 = 1
−1 −1
2 must be at most four. Hence the sum of the
dimensions of the three eigenspaces is exactly
four. . Thus
f (A) = Nul(A − λ2 I) must be at least one-dimensional. Thus the
sum of the dimensions of = 8 − (3 − λ)(3 + λ) = λ2 − 1 = 0 , 5 Consequently, the statement is
1
1 f (1)
0
0
f (−1) 1
−1 −1
2 . TRUE . Now f (1) = 3x3 − 3x2 + 3x − 2 x=1 = 1, while 3
2
f (−1) = 3x − 3x + 3x − 2 x=−1 Consequently,
1
1
0
2 1
f (A) =
−1
0 −11
1 1
13 −24
.
=
12 −23
009 010 Using the fact that
ex = 1 + x + = −11 . −1
2 10.0 points If A is a 4 × 4 matrix having distinct eigenvalues λ1 , λ2 , λ3 such that Nul(A − λ3 I) is
two-dimensionsional, then A is diagonalizable.
True or False? 10.0 points 1
1 2
x + . . . + xn + . . . ,
2!
n! compute etA as a matrix-valued function of t
when
−8 10
.
A =
−5 7
" 2t
#
−3t
−3t
2t
2e
−
e
2(e
−
e
)
1. etA =
e2t − e−3t
2e−3t − e2t
2. etA = " 2e−3t − e2t 2(e2t − e−3t ) e−3t − e2t 2e2t − e−3t correct 3. etA = " 2e2t + e2t 2(e−3t − e2t ) e2t − e−3t 2e−3t + e2t # # yick (tay236) – HW 10 – gilbert – (53415) 4. etA = " 2e−3t + e2t 2(e2t − e−3t ) e−3t − e2t 2e2t + e−3t is a diagonalization of the matrix # 3 A = −6
15 Explanation:
If A can be diagonalized by
d1 0
−1
P −1 ,
A = P DP
= P
0 d2
e tD = Pe P −1 = P etd1
0 0 etd2 P 0 0
1 0 .
5 −4 1. d2 = 1, d3 = −4, then
tA 6 −1 0 P = −3 0 . Now A can be diagonalized if we can find
an eigenbasis of R2 of eigenvectors v1 , v2 of A
corresponding to eigenvalues λ1 , λ2 , for then:
λ1 0
P −1 , P = [v1 v2 ] .
A = P
0 λ2 1
0 1 0 0 = λ2 − λ − 6 = (λ + 3)(λ − 2) = 0 , P = , P −1 Consequently,
−3t
e
2 1
tA
e =
0
1 1
= " 2e−3t − e2t
e −3t 011 2t −e = 0
e2t 1
−1 −1
2 1
−1 2e − e −3t . −1
2 # . 2(e2t − e−3t )
2t 10.0 points Find a matrix P and d2 , d3 so that 3 0 0
P 0 d2 0 P −1 , d1 ≥ d2 ≥ d3 ,
0 0 d3 0 3. d2 = 4, d3 = −1, 1 0 P = −3 0
0 so 2 1
1 1 1 P = 0 1 −3 i.e., λ1 = −3 and λ2 = 2. Corresponding
eigenvectors are
1
2
,
,
v2 =
v1 =
1
1
1 1 1 1 10 7 − λ 2. d2 = 1, d3 = −4, But −8 − λ
det[A − λI] = −5 0 0 1 1 1 0 0 4. d2 = 1, d3 = −4, 1 P = −3 1
0 1 0 1 correct
5. d2 = 4, d3 = −1, 0 0 1 P = 0 1 −3 1 1 0 yick (tay236) – HW 10 – gilbert – (53415)
6. d2 = 4, d3 = −1, 1 0 P = −3 1
0 0 0 1 1 Explanation:
The entries 3, d2 , d3 in the diagonal matrix
are the respective eigenvalues λ1 , λ2 , λ3 of A.
But 3 − λ
0
0 1−λ
0 det[A − λI] = −6 15
5
−4 − λ = −λ3 + 13λ − 12 = −(λ − 3)(λ − 1)(λ + 4) .
So λ1 = 3, λ2 = 1, λ3 = −4.
Now let u1 , u2 , u3 be eigenvectors of A corresponding to λ1 , λ2 , λ3 respectively. Since
the eigenvalues are distinct, To determine u2 we row reduce A − λI with
λ2 = 1: 2 0 0
rref(A − I) = rref −6 0 0 15 5 −5 1 0 0
= 0 1 −1 .
0 0 0 Thus 0
u2 = 1 .
1 To determine u3 we row reduce A − λI with
λ3 = −4: 7 0 0
rref(A + 4I) = rref −6 5 0 15 5 0 −1 0 0
= 0 −1 0 .
0
0 0 Thus, finally, P = [u1 u2 u3 ]
has orthogonal columns. λ1 0
A = P 0 λ2
0
0
is a diagonalization of A. 0
0 P −1
λ3 To determine u1 we row reduce A − λI with
λ1 = 3: 0
0
0
rref(A − 3I) = rref −6 −2 0 15
5 −7 −3 −1 0
= 0
0 1.
0
0 0
Thus 1
u1 = −3 .
0 7 0 u3 = 0 .
1 Consequently, d2 = 1, d3 = −4 and 1 P = −3
0 012 0 0 1 0 .
1 1 10.0 points Find a matrix P so that
d1 0
P −1 ,
P
0 d2 d1 ≥ d2 is a diagonalization of the matrix
0
0
A=
−8 −4
1
1. P =
−2
0
correct
1 yick (tay236) – HW 10 – gilbert – (53415) 2. P 3. P 4. P 5. P 6. P So, P = [u1 u2 ] and
2 1
=
1 −3
−2 1
=
1 0
−1 −3
=
−2 1
0 1
=
1 −2
0 1
=
−1 2
A = P DP −1
is a diagonalization of A.
Consequently,
0
D=
0 013 det(A − λI) = (0 − λ)(−4 − λ)
= λ2 + 4λ
= (λ + 0)(λ + 4) = 0.
0 0
λ1 0
.
=
0 −4
0 λ2
Now to find the eigenvectors of A, we will
solve for the nontrivial solution of the characteristic equation by row reducing the related
augmented matrices:
0+0
0
0
[ A − λ1 I 0 ] =
−8 −4 + 0 0
−2 −1 0
0
0 0
∼
=
0
0 0
−8 −4 0
1
,
=⇒ u1 =
−2
D= while 0+4
0
0
−8 −4 + 4 0
1 0 0
4 0 0
∼
=
0 0 0
−8 0 0
0
=⇒ u2 =
.
1 [ A − λ2 I 0] =
1
0
, P =
−2
−4 Explanation:
To begin, we must find the eigenvectors
and eigenvalues of A. To do this, we will use
the characteristic equation, det(A − λI) = 0.
That is, we will look for the zeros of the
characteristic polynomial. So 8 0
1 . 10.0 points Let A be a 2 × 2 matrix with eigenvalues 4
and 2 and corresponding eigenvectors
2
−2
.
, v2 =
v1 =
1
−4
Let {xk } be a solution of the difference equation
−4
.
xk+1 = Axk , x0 =
1
Compute x1 .
−10
1. x1 =
−4
−10
2. x1 =
4
4
3. x1 =
−10
10
4. x1 =
−4
−4
correct
5. x1 =
10
−4
6. x1 =
−10 Explanation:
To find x1 we must compute Ax0 . Now, express x0 in terms of v1 and v2 . That is, find
c1 and c2 such that x0 = c1 v1 + c2 v2 . This
is certainly possible because the eigenvectors yick (tay236) – HW 10 – gilbert – (53415)
v1 and v2 are linearly independent (by inspection and also because they correspond to
distinct eigenvalues) and hence form a basis
for R2 . The row reduction
−2 2 −4
[ v1 v2 x0 ] =
−4 1 1
1 0 −1
∼
0 1 −3
shows that x0 = −v1 − 3v2 . Since v1 and v2
are eigenvectors (for the eigenvalues 4 and 2
respectively):
x1 = Ax0 = A(−v1 − 3v2 )
= −Av1 − 3Av2 = −4v1 − 3 · 2v2
−4
−12
8
.
=
+
=
10
−6
16
Consequently,
x1 = 014 −4
10 . Let A be a 2 × 2 matrix with eigenvalues 5
and 3 and corresponding eigenvectors
4
−3
.
, v2 =
v1 =
−13
12
Determine the solution {xk } of the difference
equation
2
.
xk+1 = Axk , x0 =
−11
1. xk = −2 (5)k v1 + 2 (3)k v2
k 2. xk = −2 (5) v1 − (3) v2 correct
3. xk = −4 (5)k v1 − 2 (3)k v2
4. xk = −4 (5)k v1 − (3)k v2
5. xk = −2 (5)k v1 − 2 (3)k v2
6. xk = −4 (5)k v1 + (3)k v2 Explanation:
Since v1 and v2 are eigenvectors corresponding to distinct eigenvalues of A, they
form an eigenbasis for R2 . Thus
x0 = c1 v1 + c2 v2
To compute c1 and c2 we apply row reduction
to the augmented matrix
−3
4
2
[ v1 v2 x0 ] =
12 −13 −11
1 0 −2
.
∼
0 1 −1
This shows that c1 = −2, c2 = −1 and
x0 = −2v1 −v2 . Since v1 and v2 are eigenvectors corresponding to the eigenvalues 5 and 3
respectively,
xk = Ak x0 = Ak (−2v1 − v2 ) 10.0 points k 9 = −2Ak v1 − Ak v2
= −2 (5)k v1 − (3)k v2
for k ≥ 0. It follows that
Axk = −2 (5)k Av1 − (3)k Av2
= −2 (5)k+1 v1 − (3)k+1 v2 = xk+1 .
This shows that {xk } solves the difference
equation. Consequently,
xk = −2 (5)k v1 − (3)k v2 . 015 10.0 points Let A be a 3 × 3 matrix with eigenvalues
3, 2, and −2 and corresponding eigenvectors 2
−3
1
v1 = 6 , v2 = −12 , v3 = 2 .
4
−15
−3 If {xk } is the solution of the difference equation 0
xk+1 = Axk ,
x0 = −2 ,
−4 yick (tay236) – HW 10 – gilbert – (53415)
determine x1 . 12
1. x1 = −16 −8 8
2. x1 = −16 −12 −12
3. x1 = 16 8 8
4. x1 = 16 correct
−12 12
5. x1 = 16 −8 −8
6. x1 = −16 12 Explanation:
To find x1 we must compute Ax0 . First, we
express express x0 in terms of v1 , v2 , and v3 :
x0 = c1 v1 + c2 v2 + c3 v3 .
This is certainly possible as the eigenvectors
v1 , v2 , and v3 are linearly independent because the eigenvalues are distinct. Hence they
form a basis for R3 . The row reduction 2 −3
1
0
[ v1 v2 v3 x0 ] = 6 −12 2 −2 4 −15 −3 −4 1 0 0 2
∼ 0 1 0 1 0 0 1 −1
shows that x0 = 2v1 +v2 −v3 . But v1 , v2 and
v3 are eigenvectors for the respective eigenvalues 3, 2 and −2, so
x1 = Ax0 = A(2v1 + v2 − v3 )
= 2Av1 + Av2 − Av3
= 2 · (3)v1 + (2)v2 − (−2)v2 −6
2
8
12
= 36 + −24 + 4 = 16 .
24
−30
−6
−12 10 Consequently, 8
x1 = 16 .
−12
016 10.0 points Let A be a 3 × 3 matrix with eigenvalues 2,
−1, and −3 and corresponding eigenvectors 2
v1 = 2 ,
−6 3
v2 = 0 ,
−3 −3
v3 = −2 .
8 Determine the solution {xk } of the difference
equation xk+1 = Axk , −6
x0 = −4 .
13 1. xk = 3 (2)k v1 − (−1)k v2 − (−3)k v3
2. xk = −3 (2)k v1 + (−1)k v2 − (−3)k v3
3. xk = 3 (2)k v1 − (−1)k v2 + (−3)k v3
4. xk = −3 (2)k v1 + (−1)k v2 + (−3)k v3
5. xk = 3 (2)k v1 + (−1)k v2 + (−3)k v3
6. xk = −3 (2)k v1 − (−1)k v2 − (−3)k v3
correct
Explanation:
Since v1 , v2 , and v3 are eigenvectors corresponding to distinct eigenvalues of A, they
form an eigenbasis for R3 . Thus
x0 = c1 v1 + c2 v2 + c3 v3 yick (tay236) – HW 10 – gilbert – (53415)
To compute c1 , c2 , and c3 we apply row reduction to the augmented matrix 2
3 −3 −6
[ v1 v2 v3 x0 ] = 2
0 −2 −4 −6 −3 8
13 1 0 0 −3
∼ 0 1 0 −1 .
0 0 1 −1 This shows that c1 = −3, c2 = −1, c3 = −1
and x0 = −3v1 − v2 − v3 . Since v1 , v2 ,
and v3 are eigenvectors corresponding to the
eigenvalues 2, −1, −3 respectively,
xk = Ak x0 = Ak (−3v1 − v2 − v3 )
= −3Ak v1 − Ak v2 − Ak v3 for k ≥ 0. It follows that
Axk = −3 (2)k Av1 − (−1)k Av2 − (−3)k Av3
= −3 (2)k+1 v1 − (−1)k+1 v2 − (−3)k+1 v3
= xk+1 .
This shows that {xk } solves the difference
equation. Consequently,
xk = −3 (2)k v1 − (−1)k v2 − (−3)k v3 .
10.0 points Let A be a 3 × 3 matrix with eigenvalues 3,
3
− 1 and corresponding eigenvectors
2 , and 2 −2
−1
−3
v1 = 4 , v2 = 3 , v3 = 7 .
2
−2
−3 Determine the solution {xk } of the difference
equation −1
xk+1 = Axk , x0 = 1 .
−2
1. xk = − (3)k v1 + 3
k 2. xk = (3) v1 + 3
3 k
1 k 2 v2 + 2 − 2
3 k
2 k
− 12 v2 − 2
3 k
2 4. xk = − (3)k v1 − 3
correct
5. xk = (3)k v1 − 3
3 k
2 6. xk = − (3)k v1 + 3 v2 + 2 − 12
3 k
2 v3 v2 + 2 − 12 v2 − 2 − 12
3 k
2 k k v2 − 2 − 12 k v3 v3
k v3 Explanation:
Since v1 , v2 , and v3 are eigenvectors corresponding to distinct eigenvalues of A, they
form an eigenbasis for R3 . Thus
x0 = c1 v1 + c2 v2 + c3 v3 = −3 (2)k v1 − (−1)k v2 − (−3)k v3 017 3. xk = (3)k v1 − 3 11 v3 v3 To compute c1 , c2 , and c3 we apply row reduction to the augmented matrix −2 −1 −3 −1
[ v1 v2 v3 x0 ] = 4
3
7
1 2 −2 −3 −2 1 0 0 −1
∼ 0 1 0 −3 .
0 0 1 2 This shows that c1 = −1, c2 = −3, c3 = 2
and x0 = −v1 − 3v2 + 2v3 . Since v1 , v2 ,
and v3 are eigenvectors corresponding to the
eigenvalues 3, 32 , − 12 respectively,
xk = Ak x0 = Ak (−v1 − 3v2 + 2v3 )
= −Ak v1 − 3Ak v2 + 2Ak v3
k
k
= − (3)k v1 − 3 23 v2 + 2 − 12 v3 for k ≥ 0. It follows that
Axk = − (3)k Av1 − 3
= − (3)k+1 v1 − 3
= xk+1 .
3 k
1 k
Av
+
2
−
Av3
2
2
2
k+1
k+1
3
v2 + 2 − 21
v3
2 This shows that {xk } solves the difference
equation. Consequently,
xk = − (3)k v1 − 3
3 k
2 v2 + 2 − 12 k v3 . yick (tay236) – HW 10 – gilbert – (53415) 12 Thus
018 10.0 points Find the solution of the differential equation
du
0
= Au(t), u(0) =
−8
dt
when A is a 2 × 2 matrix with eigenvalues 3
and 2 and corresponding eigenvectors
2
2
.
, v2 =
v1 =
−10
−8
1. u(t) = −8e3t v1 + 8e2t v2
2. u(t) = −4e3t v1 + 4e2t v2 correct u(t) = −4e3t v1 + 4e2t v2
solves the given differential equation.
019 Find the solution of the differential equation
du
−1
= Au(t), u(0) =
0
dt
when A is the matrix
−13
A=
8 3. u(t) = −4e3t v1 + 8e2t v2
4. u(t) = −8e3t v1 + 4e2t v2
5. u(t) = −4e3t v1 − 8e2t v2 Explanation:
Since v1 and v2 are eigenvectors corresponding to distinct eigenvalues of A, they
form an eigenbasis for R2 . Thus
u(0) = c1 v1 + c2 v2
To compute c1 and c2 we apply row reduction
to the augmented matrix
2
2
0
[ v1 v2 u(0) ] =
−8 −10 −8
1 0 −4
.
∼
0 1 4 Since v1 and v2 are eigenvectors corresponding to the eigenvalues 3 and 2 respectively,
2t u(t) = −4e v1 + 4e v2 .
Then u(0) is the given initial value and
Au(t) = −4e3t Av1 + 4e2t Av2
du(t)
= −4 3e3t v1 + 4 2e2t v2 =
.
dt 1. u(t) = 2e−t + 3e−5t
2e−t + 2e−5t 2. u(t) = −2e−t − 3e−5t
2e−t + 2e−5t 3. u(t) = −2e−t + 3e−5t
2e−t − 2e−5t 4. u(t) = 2e−t + 3e−5t
−2e−t − 2e−5t 5. u(t) = 2e−t − 3e−5t
−2e−t + 2e−5t 6. u(t) = −2e−t − 3e−5t
−2e−t − 2e−5t This shows that c1 = −4, c2 = 4 and
u(0) = −4v1 + 4v2 . −12
7 6. u(t) = −8e3t v1 − 4e2t v2 3t 10.0 points correct Explanation:
Since −13 − λ
det[A − λI] = 8 −12 7 −λ = (−13 − λ)(7 − λ) + 96 = λ2 + 6λ + 5 = (λ + 1)(λ + 5), yick (tay236) – HW 10 – gilbert – (53415) 13 the eigenvalues of A are λ1 = −1, λ2 = −5
and corresponding eigenvectors
3
1
, v2 =
v1 =
−2
−1 Find the solution of the differential equation
du
−8
= Au(t), u(0) =
−3
dt form a basis for R2 because λ1 6= λ2 . Thus when A is the matrix
1 1 5
A=
2 0 6 u(0) = c1 v1 + c2 v2 .
To compute c1 and c2 we apply row reduction
to the augmented matrix
1
3 −1
[ v1 v2 u(0) ] =
−1 −2 0
1 0 2
.
∼
0 1 −1
This shows that c1 = 2, c2 = −1 and Then u(0) is the given initial value and
Au(t) = 2e−t Av1 − e−5t Av2
du(t)
= 2 −e−t v1 − −5e−5t v2 =
.
dt Thus u(t) is a solution of the differential equation. But
3
1
−5t
−t
+ (−)e
u(t) = 2e
−2
−1
2e−t − 3e−5t
=
.
−2e−t + 2e−5t 1 1 1. u(t) = 3e3t − 5e 2 t
−3e3t 2. u(t) = 3e3t − 5e 2 t
3e3t 3. u(t) = 3e3t + 5e 2 t
3e3t 4. u(t) = −3e3t + 5e 2 t
3e3t 5. u(t) = −3e3t + 5e 2 t
−3e3t 6. u(t) = −3e3t − 5e 2 t
−3e3t u(0) = 2v1 − v2 .
Since v1 and v2 are eigenvectors corresponding to the eigenvalues −1 and −5 respectively,
set
u(t) = 2e−t v1 − e−5t v2 . 1 1 1 1 correct Explanation:
Since 1 −λ 2
det[A − λI] = 0 5
2 3 − λ = ( 21 − λ)(3 − λ) + 0
3
7
= λ2 − λ + = (λ − 3)(λ− 12 ),
2
2 Consequently,
u(t) = 2e−t − 3e−5t
−2e−t + 2e−5t solves the given differential equation. the eigenvalues of A are λ1 = 3, λ2 =
corresponding eigenvectors
−1
−1
, v2 =
v1 =
0
−1 1
2 and form a basis for R2 because λ1 6= λ2 . Thus
020 10.0 points
u(0) = c1 v1 + c2 v2 . yick (tay236) – HW 10 – gilbert – (53415)
To compute c1 and c2 we apply row reduction
to the augmented matrix
−1 −1 −8
[ v1 v2 u(0) ] =
−1 0 −3
1 0 3
.
∼
0 1...

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