**Unformatted text preview: **berana (sb54896) – HW 07 – gilbert – (53435)
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001 10.0 points A = A is not diagonalizable . 10.0 points If the matrix
2
1 −1
4 A = 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 3
3 0
4. D =
0 2
5. A is not diagonalizable correct 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 ?
2 4
1. P =
−1 1
−1 4
correct
2. P =
1 1
−1 4
3. P =
2 1
−1 1
4. P =
1 4
5. A is not diagonalizable Explanation:
Since
det [A − λI] = Consequently, 002 If the matrix 1 2−λ
1 −1
4−λ Explanation:
Since det [A − λI] = = 1 + (2 − λ)(4 − λ) = 9 − 6λ + λ2 , 4
−2 − λ the eigenvalues of A are the solutions of 9 − 6λ + λ2 = (3 − λ)2 = 0 , λ2 + λ − 6 = (λ + 3)(λ − 2) = 0 , i.e., λ = 3, 3. 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) . 1−λ
1 = −4 − (1 − λ)(2 + λ) = λ2 + λ − 6 , the eigenvalues of A are the solutions of On the other hand, when λ = 3,
1
−1 −1
=
rref(A − λI) = rref
0
1
1 i.e., λ = −3, 2. Thus A is diagonalizable
because the eigenvalues of A are distinct, and
A = P DP −1 with
P = [v1 v2 ]
where v1 and v2 are eigenvectors corresponding to λ1 and λ2 respectively. To determine
v1 and v2 we solve the equation Ax = λx. berana (sb54896) – HW 07 – gilbert – (53435)
Ax = λx, λ = −3:
x
x1 + 4x2
x1
1 4
= −3 1 ,
=
x2
x1 − 2x2
x2
1 −2
which can be written as
x1 + 4x2 = −3x1 , x1 − 2x2 = −3x2 , Ax = λx, λ = 2:
x
x1 + 4x2
x1
1 4
= 2 1 ,
=
x2
x1 − 2x2
x2
1 −2
which can be written as
x1 − 2x2 = 2x2 , i.e., x1 = 4x2 . So one choice of v2 is
4
.
v2 =
1
Consequently, A = P DP −1 with
−1 4
P =
.
1 1
003 1 0 2
3. P = 1 1 0 0 0 1 4. A is not diagonalizable i.e., x1 = −x2 . So one choice of v1 is
−1
.
v1 =
1 x1 + 4x2 = 2x1 , 2 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 correct
0 0 1 1 0 −2
2. P = 1 1 0 0 0 1 −1 5. P =
1
0 0 2
1 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 berana (sb54896) – HW 07 – gilbert – (53435)
Thus Nul(A − 3I) has dimension 2, and −2
0
v2 = 1 , v3 = 0 ,
0
1
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
004 10.0 points 2. f (A) = 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 . f (A) = P f (D)P −1
f (d1 )
0
P −1 .
= P
0
f (d2 )
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
But 3 − λ
det[A − λI] = 2 By diagonalizing the matrix
3 −4
,
A =
2 −3
compute f (A) for the polynomial
f (x) = 3x3 − x2 − 2x − 2 . 1. f (A) = 0
−2 4
6 −4 −3 − λ = 8 − (3 − λ)(3 + λ) = λ2 − 1 = 0 , FALSE .
10.0 points 0
−2 then Consequently, the statement is 005 Explanation:
If A can be diagonalized by
d1 0
−1
P −1 ,
A = P DP
= P
0 d2 True or False? 2. TRUE 0 4
2 −6
−4
3. f (A) =
6
0 −4
correct
4. f (A) =
2 −6
Every n × n matrix A having eigenvectors
v1 , v2 , . . . , vn can be diagonalized. 1. FALSE correct 3 i.e., λ1 = 1 and λ2 = −1. Corresponding
eigenvectors are
1
2
,
,
v2 =
v1 =
1
1
so
P = 2
1 2 1
1 1 1
1 , P −1 = 1
−1 −1
2 . Thus
f (A) = f (1)
0
0
f (−1) 1
−1 −1
2 . berana (sb54896) – HW 07 – gilbert – (53435)
Now f (1) = 3x3 − x2 − 2x − 2 x=1 = −2 , while f (−1) = 3x − x − 2x − 2 3 2 = −4 . x=−1 −1
2 10.0 points
1 2
1
x + . . . + xn + . . . ,
2!
n! tA compute e
when 1. etA = as a matrix-valued function of t
−7 8
.
A =
−4 5
#
" t
2e + et 2(e−3t − et )
et − e−3t
t 2. etA = " 2e − e 3. etA = " 2e−3t + et tA 4. e = t 2e−3t + et
2(e −3t −3t e −e " 2e −3t e −3t −3t t −e )
t −e 2(et − e−3t ) t t −e 2e + e −3t # 2(et − e−3t ) e−3t − et 2et − e−3t # cor- e = Pe P = P etd1
0 0 etd2 P P = 2
1 1
1 , −1 P −1 = 0
et Consequently,
−3t
2 1
e
tA
e =
0
1 1
" 1
−1 . . 1
−1 −1
2 2(et − e−3t ) e−3t − et 2et − e−3t 10.0 points Find a matrix P so that
d1 0
P −1 ,
P
0 d2 1
4 1
−4 0
2. P =
1 −1
2 2e−3t − et −4
1. P =
1 then
−1 so d1 ≥ d2 is a diagonalization of the matrix
2 8
A=
0 0 Explanation:
If A can be diagonalized by
d1 0
−1
P −1 ,
A = P DP
= P
0 d2
tD i.e., λ1 = −3 and λ2 = 1. Corresponding
eigenvectors are
1
2
,
,
v2 =
v1 =
1
1 007 rect tA = λ2 − λ − 3 = (λ + 3)(λ − 1) = 0 , = # 2e−3t − et −7 − λ
8 det[A − λI] = −4
5 − λ Using the fact that
ex = 1 + x + 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
But Consequently,
1
−2 0
2 1
f (A) =
−1
0 −4
1 1
0 −4
.
=
2 −6
006 4 # . berana (sb54896) – HW 07 – gilbert – (53435)
1
−4 4
1 1
0 −4
3. P =
−1
−1
4. P =
0
−4
5. P =
1
1
6. P =
0 So, P = [u1 u2 ] and A = P DP −1
is a diagonalization of A.
Consequently,
1
2 0
, P =
D=
0
0 0
−4
correct
1 008 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. λ1
D=
0 0
λ2 2
=
0
0
.
0 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:
2−2
8
0
[ A − λ1 I 0 ] =
0
0−2 0
0 −1 0
0 8 0
∼
=
0 0 0
0 −2 0
1
,
=⇒ u1 =
0
while
[ A − λ2 I
= 0]
2 8
0 0
=⇒ 2+0
8
0
=
0
0+0 0
1 4 0
0
∼
0 0 0
0
−4
.
u2 =
1 . 10.0 points The eigenvalues of an n × n matrix A are
the entries on the main diagonal of A.
True or False?
1. TRUE Explanation:
The eigenvalues of a triangular n × n matrix A are the entries on the main diagonal of
A. This is not necessarily true for matrices
that are not triangular. = λ2 − 2λ
= (λ − 2)(λ + 0) = 0.
−4
1 2. FALSE correct det(A − λI) = (2 − λ)(0 − λ) So 5 Consequently, the statement is
FALSE . 009 10.0 points When
−5
A=
−1 0
−4 find matrices D and P in a diagonalization of
A given that λ1 > λ2 .
−3 −1
−4 0
, P =
1. D =
−1 1
0 −5
−4
2. D =
0
−1
0
, P =
0
−5
0
5 0
, P =
3. D =
−1
0 4
1
1 1
1
berana (sb54896) – HW 07 – gilbert – (53435)
4. D =
rect −4
0
0
0
, P =
−1
−5
−3
5 0
, P =
5. D =
−1
0 4 −1
1
−1
5 0
, P =
6. D =
0
0 4 1
1 1
1 6 So, P = [u1 u2 ] and
cor- A = P DP −1
is a diagonalization of A.
Consequently,
0
−4 0
, P =
D=
−1
0 −5 010 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.
det(A − λI) = (−5 − λ)(−4 − λ) 1
1 . 10.0 points For n × n matrix A is said to be diagonalizable when A = P DP −1 for some matrix D
and invertible matrix P .
True or False?
1. TRUE
2. FALSE correct
Explanation:
An n × n matrix A is said to be diagonalizable when A can be factored as = λ2 + 9λ + 20
= (λ + 4)(λ + 5) = 0. A = P DP −1 So λ1
D=
0 0
λ2 −4
=
0
0
.
−5 with D a DIAGONAL matrix and P an invertible matrix. 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:
−5 + 4
0
0
[ A − λ1 I 0 ] =
−1
−4 + 4 0
−1 0 0
−1 0 0
∼
=
0 0 0
−1 0 0
0
,
=⇒ u1 =
−1
while
−5 + 5
0
0
−1
−4 + 5 0
1 −1 0
0 0 0
∼
=
0 0 0
−1 1 0
1
.
=⇒ u2 =
1 [ A − λ2 I 0] = Consequently, the statement is
FALSE .
011 10.0 points If an n × n matrix A is diagonalizable, then
A has n distinct eigenvalues.
True or False?
1. TRUE
2. FALSE correct
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 3
−5
3 3
−3 ,
1 berana (sb54896) – HW 07 – gilbert – (53435)
then then A need not be diagonalizable. det[A − λI] = −(λ − 1)(λ + 2)2 = 0.
Thus the eigenvalues of A are λ = 1, −2, −2.
Now
1
rref(A − I) = 0
0 0 −1
1 1 ,
0 0 1 Nul(A − I) = s −1 : s in R . 1 On the other hand, so True or False?
1. FALSE
2. TRUE correct so 7 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 .
1
0
1 Since the eigenvalue λ = −2 is repeated, however, A does not have distinct eigenvalues.
Consequently, the statement is
FALSE .
012 10.0 points If A is a 7 × 7 matrix having eigenvalues
λ1 , λ2 , and λ3 such that
(i) the eigenspace corresponding to λ1 is
two-dimensional,
(ii) the eigenspace corresponding to λ2 is
three-dimensional, Explanation:
The eigenspace corresponding to λ3 must
be at least one-dimensional because all
eigenspaces must have a dimension at least
one. But if this eigenspace is one-dimensional,
then the sum of the dimensions of the three
eigenspaces is six, which is less than seven, in
which case A will not be diagonalizable.
On the other hand, if the eigenspace corresponding to λ3 is two-dimensional, then the
sum of the dimensions of the three eigenspaces
is seven, in which case A will be diagonalizable.
Consequently, the statement is
TRUE . 013 10.0 points Find the solution of the differential equation
du
5
= Au(t), u(0) =
−2
dt
when A is a 2 × 2 matrix with eigenvalues 5
and 4 and corresponding eigenvectors
3
−1
.
, v2 =
v1 =
−2
1
1. u(t) = 8e5t v1 + 3e4t v2
2. u(t) = 8e5t v1 − 3e4t v2
3. u(t) = 4e5t v1 − 6e4t v2
4. u(t) = 4e5t v1 + 3e4t v2 correct
5. u(t) = 8e5t v1 + 6e4t v2
6. u(t) = 4e5t v1 + 6e4t v2 berana (sb54896) – HW 07 – gilbert – (53435)
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
−1 3
5
[ v1 v2 u(0) ] =
1 −2 −2
1 0 4
.
∼
0 1 3
This shows that c1 = 4, c2 = 3 and
u(0) = 4v1 + 3v2 .
Since v1 and v2 are eigenvectors corresponding to the eigenvalues 5 and 4 respectively,
u(t) = 4e5t v1 + 3e4t v2 .
Then u(0) is the given initial value and
Au(t) = 4e5t Av1 + 3e4t Av2
du(t)
= 4 5e5t v1 + 3 4e4t v2 =
.
dt Thus u(t) = 4e5t v1 + 3e4t v2
solves the given differential equation.
014 10.0 points Find the solution of the differential equation
du
13
= Au(t), u(0) =
−14
dt
when A is the matrix
5
A=
−6
1. u(t) = 3
−4 12e2t − e−t
−12e2t + 2e−t 8 2. u(t) = 12e2t − e−t
12e2t − 2e−t 3. u(t) = −12e2t + e−t
−12e2t + 2e−t 4. u(t) = −12e2t + e−t
12e2t − 2e−t 5. u(t) = 12e2t + e−t
−12e2t − 2e−t 6. u(t) = −12e2t − e−t
12e2t + 2e−t
correct Explanation:
Since 5 − λ
3 det[A − λI] = −6
−4 − λ = (5 − λ)(−4 − λ) + 18 = λ2 − λ − 2 = (λ − 2)(λ + 1), the eigenvalues of A are λ1 = 2, λ2 = −1 and
corresponding eigenvectors
−1
−3
, v2 =
v1 =
2
3
form a basis for R2 because λ1 6= λ2 . Thus
u(0) = c1 v1 + c2 v2 .
To compute c1 and c2 we apply row reduction
to the augmented matrix
−3 −1 13
[ v1 v2 u(0) ] =
3
2 −14
1 0 −4
.
∼
0 1 −1
This shows that c1 = −4, c2 = −1 and
u(0) = −4v1 − v2 .
Since v1 and v2 are eigenvectors corresponding to the eigenvalues 2 and −1 respectively,
set
u(t) = −4e2t v1 − e−t v2 . berana (sb54896) – HW 07 – gilbert – (53435)
Then u(0) is the given initial value and
6. u(t) = Au(t) = −4e2t Av1 − e−t Av2
du(t)
= −4 2e2t v1 − −e−t v2 =
.
dt Thus u(t) is a solution of the differential equation. But
−t −1
2t −3
+ (−)e
u(t) = −4e
2
3
12e2t + e−t
=
.
−12e2t − 2e−t
Consequently,
u(t) = 12e2t + e−t
−12e2t − 2e−t solves the given differential equation.
015 10.0 points Find the solution of the differential equation
du
−10
= Au(t), u(0) =
−12
dt
when A is the matrix
1 −5
A=
3 0 2
−3
5 5 1. u(t) = 12e−t + 2e− 3 t
12e−t 2. u(t) = 12e−t + 2e− 3 t
−12e−t 3. u(t) = 12e−t − 2e− 3 t
12e−t 4. u(t) = −12e − 2e
12e−t 5. u(t) = −12e−t − 2e− 3 t
−12e−t 5 −t 5 5 −12e−t + 2e− 3 t
−12e−t correct Explanation:
Since 5 − − λ 3
det[A − λI] = 0 2
3 −1 − λ = (− 53 − λ)(−1 − λ) + 0
5
8
= λ2 + λ + = (λ + 1)(λ+ 53 ),
3
3 the eigenvalues of A are λ1 = −1, λ2 = − 53
and corresponding eigenvectors
−1
−3
, v2 =
v1 =
0
−3
form a basis for R2 because λ1 6= λ2 . Thus
u(0) = c1 v1 + c2 v2 .
To compute c1 and c2 we apply row reduction
to the augmented matrix
−3 −1 −10
[ v1 v2 u(0) ] =
−3 0 −12
1 0 4
.
∼
0 1 −2
This shows that c1 = 4, c2 = −2 and
u(0) = 4v1 − 2v2 .
Since v1 and v2 are eigenvectors corresponding to the eigenvalues −1 and − 35 respectively,
set
5
u(t) = 4e−t v1 − 2e− 3 t v2 .
Then u(0) is the given initial value and
5 Au(t) = 4e−t Av1 − 2e− 3 t Av2
du(t)
−t
5 − 53 t
v2 =
= 4 −e
v1 − 2 − 3 e
dt − 35 t 9
Thusu(t) solves the differential equation. But
− 53 t −1
−t −3
− 2e
u(t) = 4e
−3
0
−t
− 35 t
−12e
+
2e
=
−12e−t berana (sb54896) – HW 07 – gilbert – (53435) 10 This shows that c1 = −5, c2 = −1 and Consequently,
u(t) = 5 −12e−t + 2e− 3 t
−12e−t solves the given differential equation.
016 10.0 points Since v1 and v2 are eigenvectors corresponding to the eigenvalues 4 and 2 respectively,
set
u(t) = −5e4t v1 − e2t v2 .
Then u(0) is the given initial value and Let u(t) satisfy
du
= Au(t),
dt u(0) = −5v1 − v2 . u(0) =
6
.
−2 Compute u(2) when A is a 2 × 2 matrix with
eigenvalues 4 and 2 and corresponding eigenvectors
4
−2
.
, v2 =
v1 =
−8
2
10e8 + 4e4
1. u(2) =
10e8 + 8e4
2. u(2) =
−10e8 + 4e4
10e8 − 8e4 3. u(2) = −10e8 − 4e4
10e8 + 8e4 4. u(2) = 10e8 + 4e4
−10e8 − 8e4 5. u(2) = 10e8 − 4e4
−10e8 + 8e4 6. u(2) = −10e8 − 4e4
−10e8 − 8e4 Au(t) = −5e4t Av1 − e2t Av2
du(t)
= −5 4e4t v1 − 2e2t v2 =
.
dt Thus u(t) = −5e4t v1 − e2t v2
solves the given differential equation. Consequently,
u(2) = 017 correct 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 4
6
[ v1 v2 u(0) ] =
2 −8 −2
1 0 −5
.
∼
0 1 −1 10e8 − 4e4
−10e8 + 8e4 . 10.0 points Find the solution of the differential equation
du
−10
= Au(t), u(0) =
6
dt
when A is the matrix
−7
A=
9 −12
14 1. u(t) = −6e5t − 16e2t
−6e5t − 12e2t 2. u(t) = 6e5t + 16e2t
−6e5t − 12e2t 3. u(t) = −6e5t − 16e2t
6e5t + 12e2t 4. u(t) = 6e5t + 16e2t
6e5t + 12e2t berana (sb54896) – HW 07 – gilbert – (53435)
5. u(t) = −6e5t + 16e2t
6e5t − 12e2t 6. u(t) = 6e5t − 16e2t
−6e5t + 12e2t 11 Consequently, u(t) =
correct 6e5t − 16e2t
−6e5t + 12e2t solves the given differential equation. Explanation:
Since −7 − λ
det[A − λI] = 9 −12 14 − λ = (−7 − λ)(14 − λ) + 108 = λ2 − 7λ + 10 = (λ − 5)(λ−2), the eigenvalues of A are λ1 = 5, λ2 = 2 and
corresponding eigenvectors
4
−2
, v2 =
v1 =
−3
2
2 form a basis for R because λ1 6= λ2 . Thus
u(0) = c1 v1 + c2 v2 .
To compute c1 and c2 we apply row reduction
to the augmented matrix
−2 4 −10
[ v1 v2 u(0) ] =
2 −3
6
1 0 −3
.
∼
0 1 −4
This shows that c1 = −3, c2 = −4 and
u(0) = −3v1 − 4v2 .
Since v1 and v2 are eigenvectors corresponding to the eigenvalues 5 and 2 respectively,
set
u(t) = −3e5t v1 − 4e2t v2 .
Then u(0) is the given initial value and
Au(t) = −3e5t Av1 − 4e2t Av2
du(t)
= −3 5e5t v1 − 4 2e2t v2 =
.
dt
This shows that u(t) solves the differential
equation. But
4
2t
5t −2
− 4e
u(t) = −3e
−3
2
6e5t − 16e2t
=
.
−6e5t + 12e2t 018 10.0 points Let A be a 2 × 2 matrix with eigenvalues
−1 and − 34 and corresponding eigenvectors
3
1
.
, v2 =
v1 =
0
1
Let {xk } be a solution of the difference equation
xk+1 = Axk ,
−10
.
x0 =
−1
Compute x1 .
−1
1. x1 =
13
13
2. x1 =
−1
1
3. x1 =
13
−13
4. x1 =
−1
−1
5. x1 =
−13
13
correct
6. x1 =
1
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
v1 and v2 are linearly independent (by inspection and also because they correspond to berana (sb54896) – HW 07 – gilbert – (53435)
distinct eigenvalues) and hence
for R2 . The row reduction
1 3
[ v1 v2 x0 ] =
1 0
1 0
∼
0 1 form a basis
−10
−1
−1
−3 shows that x0 = −v1 − 3v2 . Since v1 and v2
are eigenvectors (for the eigenvalues −1 and
− 34 respectively):
x1 = Ax0 = A(−v1 − 3v2 )
=
= −Av1 − 3Av2 = −−1v1 − 3 · − 34 v2
13
12
1
.
=
+
1
0
1 Consequently,
x1 = 019 13
1 . 10.0 points Let A be a 3 × 3 matrix with eigenvalues
1, −1, and −2 and corresponding eigenvectors 1
−3
−1
v1 = −1 , v2 = 2 , v3 = 2 .
−3
6
8 If {xk } is the solution of the difference equation 2
xk+1 = Axk ,
x0 = −5 ,
−19 determine x1 . −17
1. x1 = −3 −2 2
2. x1 = −3 17 −2
3. x1 = −3 −17 12 2
4. x1 = 3 correct
17 −17
5. x1 = 3
−2 17
6. x1 = 3 2 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 1 −3 −1
2
[ v1 v2 v3 x0 ] = −1 2
2
−5 −3 6
8 −19 1 0 0 3
∼ 0 1 0 1 0 0 1 −2
shows that x0 = 3v1 + v2 − 2v3 . But v1 , v2
and v3 are eigenvectors for the respective
eigenvalues 1, −1 and −2, so
x1 = Ax0 = A(3v1 + v2 − 2v3 )
= 3Av1 + Av2 − 2Av3
= 3 · (1)v1 + (−1)v2 − 2 · (−2)v2 3
3
−4
2 = −3 + −2 + 8 = 3 .
−9
−6
32
17 Consequently, 2
x1 = 3 .
17 ...

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