**Unformatted text preview: **Version 101 – EXAM03 – gilbert – (54620)
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001 002 10.0 points The matrix 2 1 3
A = 0 4 −2 1 0 2 10.0 points Simplify the expression
(2u + 4v) · (4u − v) − k2u + vk2 is invertible.
True or False? for vectors u, v in Rn . 1. TRUE correct 1. −4kuk2 − 22 u · v − 5kvk2 2. FALSE 2. 4kuk2 + 10 u · v + 5kvk2
2 1 3. −4kuk − 10 u · v − 5kvk Explanation:
The matrix A will be invertible if and only
if det(A) 6= 0. Now 2 1 3 det(A) = 0 4 −2 1 0 2 0 4 4 −2 0 −2 +3 − = 2 1 0 0 2 1 2 2 4. −4kuk2 − 22 u · v + 5kvk2
5. 4kuk2 + 10 u · v − 5kvk2 correct
6. 4kuk2 − 22 u · v + 5kvk2
Explanation:
By linearity, = 2× 4−2− 3×4 = 2. (2u + 4v) · (4u − v) Consequently, the statement is = 2u · (4u − v) + 4v · (4u − v) TRUE . = 8u · u − 2u · v + 16v · u − 4v · v ,
while 003 10.0 points The vectors u1 and u2 shown in k2u + vk2 = (2u + v) · (2u + v) x2 = 2u · (2u + v) + v · (2u + v)
= 4u · u + 2u · v + 2v · u + v · v . 2 u2 But
v · u = u · v , u · u = kuk2 , v · v = kvk2 .
So after expansion the expression becomes
2 8kuk + 14 u · v − 4kvk = x1
2 2 − (4kuk2 + 4u · v + kvk2 )
2 u1 4kuk + 10 u · v − 5kvk 2 . are eigenvectors corresponding to eigenvalues
λ1 = −3 and λ2 = 1 respectively for a 2 × 2
matrix A. Version 101 – EXAM03 – gilbert – (54620)
Which of the following graphs contains the
vector A(u1 + u2 )? 2 x2
5. correct
2 −2 1. −2 2 x1 2 x1 x2
6.
2 x2
2. 2
x1
2
x2
3. Explanation:
As the graph of u1 , u2 shows,
1
2
.
u1 =
u1 =
2
1
But then
A(u1 + u2 ) = A u1 + A u2 = λ1 u1 + λ2 u2
−5
1
−6
.
=
+
= −3u1 + u2 =
−1
2
−3 2
x1
2 4. Consequently, A(u1 + u2 ) is contained in
the graph
x2 −2
−2 2 2 x1 Version 101 – EXAM03 – gilbert – (54620)
004 10.0 points A subset H of a vector space V is a subspace
of V if the following conditions are satisfied: The plane in R3 spanned by u1 , u2 is the
subspace W = Span{u1 , u2 }, and each y in
R3 has a unique orthogonal decomposition
y = projW y + (y − projW y) i) the zero vector of V is in H,
ii) the sum u + v is in H for all u, v in H,
iii) the scalar multiple cu is in H for all
scalars c and u in H. 3 where y − projW y is in the orthogonal complement W ⊥ . But then
dist(y, W ) = ky − projW yk. True or False?
1. TRUE correct
2. FALSE
Explanation:
Property (ii) says that H is closed under
vector addition, while property (iii) says that
H is closed under scalar multiplication. Thus
H is a subspace of V .
Consequently, the statement is
TRUE .
005 Now u1 , u2 are non-zero othogonal vectors,
so form a basis for W such that
y · u1
y · u2
projW y =
u1 +
u2 .
ku1 k2
ku2 k2
But when 3
−2
1
y = −6 , u1 = 2 , u2 = 2 ,
−9
1
−2
we see that
10.0 points Find the distance from 3
y = −6 −9
to the plane in R3 spanned by −2
1
u1 = 2 , u2 = 2 .
1
−2
1. dist = 7
2. dist = 4 −2
u1 = −3 2 ,
1 while
y · u2
ku2 k2 1
u2 = 2 .
−2 Consequently,
projW −2
1
7
y = −3 2 + 2 = −4 ,
1
−2
−5 and so
y − projW 3. dist = 6 correct
4. dist = 5 y · u1
ku1 k2 3
7
2
= −6 − −4 = −2 1 .
−9
−5
2 Thus
dist(y, W ) = 6 . 5. dist = 8
Explanation: 006 10.0 points Version 101 – EXAM03 – gilbert – (54620)
Use the fact that 1 A = −1
5 −4 9 −7
2 −4 1 −6 10
7 1 0 −1 5
∼ 0 −2 5 −6 0 0
0
0 to determine an orthogonal basis for Col(A). 1
1
1. −1 , −4 5
−1 1
2. −1 ,
5 −4
3. 2 ,
−6 −4
4. 2 ,
−6 thogonal basis: set u1 = a1 and
a2 · u 1
u1
u 2 = a2 −
ku1 k2 −4
1
(−36) = 2 −
−1 27
−6
5 −4
4/3
−8/3
= 2 + −4/3 = 2/3 .
−6
20/3
2/3 Consequently, the set of vectors −4 1 −1 , 1 5
1 −4 1 correct
1 is an orthogonal basis for Col(A). 1 −1 −1 If eigenvectors of an n × n matrix A are a
basis for Rn , then A is diagonalizable.
True or False? 007 2. FALSE 1 0 −1 5
A ∼ 0 −2 5 −6 0 0
0
0 1 0
−1 5
∼ 0 1 −5/2 3 .
0 0
0
0 1 a1 = −1 ,
5 Explanation:
An n × n matrix A is diagonalizable if and
only if A has n linearly independent eigenvectors. On the other hand, n vectors in Rn are
a basis for Rn if and only if the vectors are
linearly independent.
So if eigenvectors of A form a basis for Rn ,
they must be linearly independent, in which
case A will be diagonalizable.
Consequently, the statement is
TRUE .
008 Thus the pivot columns of A are 10.0 points 1. TRUE correct 1 −1 5 Explanation:
The pivot columns of A provide a basis for
Col(A). But by row reduction, 4 −4 a2 =
2 .
−6 We apply Gram-Schmidt to produce an or- 10.0 points Given vectors 0
4
1
u1 = −4 , u2 = −2 , u3 = 1 ,
4
−2
1 in R3 , determine c1 so that y = c1 u1 + c2 u2 + c3 u3 Version 101 – EXAM03 – gilbert – (54620)
5
when
−2
−2 .
, v2 =
v1 =
0
−5
−2
y = −1 .
Let {xk } be a solution of the difference equa3
tion
0
1. No such value of c1 exists.
.
xk+1 = Axk , x0 =
−6
1
2. c1 =
Compute x1 .
4
4
3. c1 = 0
1. x1 =
−8
1
8
4. c1 =
correct
2. x1 =
2
4
1
5. c1 = −
−4
2
3. x1 =
8
1
6. c1 = −
4
4
4. x1 =
8
Explanation:
8
Since
correct
5. x1 =
−4
u1 · u2 = u1 · u3 = u2 · u3 = 0,
−8
6. x1 =
4
the vectors u1 , u2, u3 are mutually orthogo3
nal, hence form a basis for R as they are also
Explanation:
non-zero. Thus by orthogonality,
To find x1 we must compute Ax0 . Now, exy = proju1 y + proju2 y + proju3 y
press x0 in terms of v1 and v2 . That is, find
c1 and c2 such that x0 = c1 v1 + c2 v2 . This
y · u2
y · u3
y · u1
u1 +
u2 +
u3 ,
=
is certainly possible because the eigenvectors
u1 · u1
u2 · u2
u3 · u3
v1 and v2 are linearly independent (by inso
spection and also because they correspond to
y · u1
y · u2
y · u3
distinct eigenvalues) and hence form a basis
c1 =
, c2 =
, c3 =
.
for R2 . The row reduction
u1 · u1
u2 · u2
u3 · u3
But
−2 −2 0
[ v1 v2 x0 ] =
−2 −5 −6
y · u1
(0) + (4) + (12)
=
.
1 0 −2
u1 · u1
(0) + (16) + (16)
∼
0 1 2
Consequently,
shows that x0 = −2v1 + 2v2 . Since v1 and v2
are eigenvectors (for the eigenvalues 4 and 2
1
c1 =
.
respectively):
2
009 10.0 points Let A be a 2 × 2 matrix with eigenvalues 4
and 2 and corresponding eigenvectors x1 = Ax0 = A(−2v1 + 2v2 )
= −2Av1 + 2Av2 = −2 · 4v1 + 2 · 2v2
8
−8
16
.
=
+
=
−4
−20
16 Version 101 – EXAM03 – gilbert – (54620) 6 This shows that c1 = −1, c2 = 4 and Consequently,
x1 = 010 8
−4 u(0) = −v1 + 4v2 . . Since v1 and v2 are eigenvectors corresponding to the eigenvalues 4 and 2 respectively,
set
u(t) = −e4t v1 + 4e2t v2 . 10.0 points Let u(t) satisfy
du
= Au(t),
dt u(0) =
−13
.
14 Compute u(2) when A is a 2 × 2 matrix with
eigenvalues 4 and 2 and corresponding eigenvectors
−3
1
.
, v2 =
v1 =
3
−2
−e8 + 12e4
1. u(2) =
−2e8 + 12e4
2. u(2) = −e8 + 12e4
2e8 − 12e4 3. u(2) = e8 − 12e4
2e8 − 12e4 4. u(2) = e − 12e
−2e8 + 12e4 5. u(2) = e8 + 12e4
−2e8 − 12e4 6. u(2) = −e8 − 12e4
2e8 + 12e4 Then u(0) is the given initial value and
Au(t) = −e4t Av1 + 4e2t Av2
du(t)
= − 4e4t v1 + 4 2e2t v2 =
.
dt
Thus
u(t) = −e4t v1 + 4e2t v2
solves the given differential equation. Consequently,
u(2) =
011 8 4 correct −e8 − 12e4
2e8 + 12e4 . 10.0 points If {x1 , x2 , x3 } is a linearly independent set
and
W = Span{x1 , x2 , x3 },
then any orthogonal set {v1 , v2 , v3 } in W is
a basis for W .
True or False?
1. TRUE 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 −13
[ v1 v2 u(0) ] =
−2 3
14
1 0 −1
.
∼
0 1 4 2. FALSE correct
Explanation:
The three orthogonal vectors must be nonzero to be a basis for a three-dimensional
subspace.
Consequently, the statement is incomplete
and thus
FALSE .
012 10.0 points Version 101 – EXAM03 – gilbert – (54620)
By diagonalizing the matrix
3
4
,
A =
−2 −3
compute f (A) for the polynomial
3 2 f (x) = 4x + x − 3x + 1 .
−5
2 −4
1 2. f (A) = 5
−2 3. f (A) = 5
2
4
correct
−1
4. f (A) = −5
−2 1. f (A) = 4
1 −4
−1 so
P = 2
−1 1
−1 −1 = 1
−1 , P f (1)
0
0
f (−1) 1
−2 Explanation:
If A can be diagonalized by
d1 0
−1
P −1 ,
A = P DP
= P
0 d2
then
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 4 −3 − λ = 8 − (3 − λ)(3 + λ) = λ2 − 1 = 0 , i.e., λ1 = 1 and λ2 = −1. Corresponding
eigenvectors are
1
2
,
,
v2 =
v1 =
−1
−1 . Thus
f (A) = 2
−1 1
−1 1
−1 1
−2 Now 3 − λ
det[A − λI] = −2 7 f (1) = 4x3 + x2 − 3x + 1 x=1 = 3, while f (−1) = 4x3 + x2 − 3x + 1 x=−1 Consequently,
1
3 0
2
1
f (A) =
−1
0 1
−1 −1
5
4
.
=
−2 −1 = 1. 1
−2 . ...

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