Name
Math 311
Final Exam Version A
Section 502
1-8
/40
10
/20
9
/16
11
/28
Total
/104
Spring 2015
Solutions
P. Yasskin
Multiple Choice: 5 points each. No Part Credit
Work Out: Points indicated. Show all work.
1. Hydrocloric acid H Cl and sodium hydroxide NaOH react to produce sodium chloride Na Cl and
water H 2 O according to the chemical equation:
aH Cl bNa O H
cNa Cl
dH 2 O
Which of the following is the augmented matrix which is used to solve this chemical equation?
(Put the elements in the order H, Cl, Na, O.)
a.
c.
1
1
0
0
0
1
0
1
1
0
0
1
1
0
0
2
0
0
1
0
1 1 0 0
0
1 0 1 1
0
0 1 1 0
0
2 0 0 1
0
d.
H:
Solution:
b.
a
b
2d
Cl :
a
c
Na :
b
c
O:
b
d
2. Suppose A is nilpotent, i.e. A 2
a.
b.
c.
d.
e.
A
A
A
A
A
1
1
1
1
1
Solution:
is invertible and
is invertible and
is invertible and
is invertible and
is not invertible
A
1 1
A
A
A
A
A
A
0
1 0 1 0
0
0 1 1 0
0
0 1 0 1
0
1 1
0
2
0
1 0
1
0
0
0 1
1
0
0
0 1
0
1
0
Correct Answer
Each equation gets a row.
0. Which of the following is true?
1
1
1
1
1
1 1 0 2
1
1
1
1
A2
A 1
1 2A
1 A
1 A
A
1
Correct Answer
A2
1.
1
3. Consider the vector space M 2, 2
1 0
E1
of 2
E2
0 1
1
0
0
1
2 matrices with the basis
0 1
E3
1 0
9 5
Which of the following are the components of the matrix A
a.
A
E
2 1 4 3
b.
A
E
5 4 3 2
c.
A
E
4 5 2 3
d.
A
E
1 2 3 4
e.
A
E
5
2E 1
1E 2
4E 3
3E 4
A
4E 1
5E 2
2E 3
3E 4
A
5E 1
4E 2
3E 3
2E 4
3 7
1 1
9
5
1
1
A
5E 1
4E 2
3E 3
2E 4
A
1E 1
2E 2
3E 3
4E 4
of real valued function on the interval
2
spanned by the basis v 1
subspace V Span x, x
orthonormal basis for V?
5 x
2
x
u2
c. u 1
d. u 1
e. u 1
3 x
2
2x
x2, x
w2, w2
x2, x2
7
1
1
1
x, v 2
whose second
f x g x dx. Consider the
f, g
2
0, 1
1
x . Which of the following is an
7 x2
2
3
2
x
x
4
5 x2
u2
Correct Answer
2
u2 4 5 x2 3 5 x
w1, w1
v2, w1
3
u2
1
x
1 1
x2
We apply Gram-Schmidt to v 1
w1
9 5
5 9
derivatives exist and are continuous with the inner product
b. u 1
relative to the E basis?
1 1
1 1
4. Consider the vector space C 2 0, 1
u2
1 0
2
A
x
1
Correct Answer
4 3
a. u 1
0
E4
x, x
x 2 dx
2
3
w2
v2
1
1
x 3 dx
1
1
x 4 dx
1
x2
x, v 2
0
2
5
u2
u1
3 x
2
x2
5 x2
2
2
5. Consider the second derivative linear operator L : P 6
P6 : L p
polynomials of degree less than 6. Find the image, Im L .
HINT: Let p a bx cx 2 dx 3 ex 4 fx 5 .
a.
b.
c.
d.
e.
Im
Im
Im
Im
Im
L
L
L
L
L
Im L
Span
Span
Span
Span
Span
Lp
2c
6dx
12ex 2
20fx 3
L
L
L
L
L
Span
Span
Span
Span
Span
P6 : L p
on the space of
1
1, x
Correct Answer
1, x, x 2
x2, x3, x4, x5
x, x 2 , x 3 , x 4 , x 5
Ker L
p Lp
0
2
Lp
2c 6dx 12ex 20fx 3 0
Ker L
p a bx
Span 1, x
7.
d2p
dx 2
Span 1, x, x 2 , x 3
polynomials of degree less than 6. Find the kernel, Ker L .
HINT: Let p a bx cx 2 dx 3 ex 4 fx 5 .
Ker
Ker
Ker
Ker
Ker
on the space of
1
1, x
1, x, x 2 , x 3
Correct Answer
2 3 4 5
x ,x ,x ,x
x, x 2 , x 3 , x 4 , x 5
6. Consider the second derivative linear operator L : P 6
a.
b.
c.
d.
e.
d2p
dx 2
Compute the line integral
c
d
e
F ds clockwise around the
f
0
y
3
complete boundary of the plus sign, shown at the right,
for the vector field
a.
25
b.
5
F
2y, 4y 3
3x .
2
1
0
c. 0
0
d. 5
e. 25
4x 3
1
2
3
x
Correct Answer
We let P, Q
F
4x 3 2y, 4y 3 3x and apply Green’s Theorem.
There is an extra minus sign because we need F ds clockwise, but Green’s Theorem expects
counterclockwise:
Q
P dx dy
F ds
P dx Q dy
3 2 dx dy 5 Area 5 5 25
x
y
3
8. Consider the vector space M 2, 2
of 2
function, L : M 2, 2
M 2, 2 : L X
AX
corresponding eigenmatrix (eigenvector) of L?
x y
HINT: Let X
1 0
2 matrices. Let A
. Consider the linear
0 2
XA. Which of the following is not an eigenvalue and
.
z w
Solution:
LX
a.
2
X
b.
1
X
c.
0
X
d.
0
X
1
e.
9.
L
x y
1 0
x y
x y
1 0
0
y
z w
0 2
z w
z w
0 2
z
0
0 1
1 0
Correct Answer
L
0 0
L
1 0
1 0
L
0 0
0 0
L
0 1
0 1
X
L
0 0
16 points
Which of the following is an inner product on
Put ’s in the correct boxes. No part credit.
Let x
x1, x2 , y
y1, y2 .
0 1
0
1
1 0
1
0
0 0
0 0
1 0
1 0
1 0
0 0
0 0
0 0
0 0
0 0
0 1
0 0
0 1
0
1
0 0
0
0
2
2
1
0
0
1
0 1
1 0
0 0
1 0
1 0
0 0
0 0
0 1
0 1
0 0
? If not, why not?
Why not?
Inner Product?
Yes
x, y
a. x 1 y 1
2x 2 y 2
b. x 1 y 1
2x 1 y 2
c. x 21 y 21
2x 22 y 22
d. x 1 y 1
x1y2
x2y1
2x 2 y 2
e. x 1 y 1
x1y2
x2y1
2x 2 y 2
f.
x1y1
x1y2
x2y1
x2y2
g. x 1 y 1
x1y2
x2y1
x2y2
h. x 1 y 1
x2y2
No
Not
Not
Not
Positive but Not
Symmetric Linear Positive Positive Definite
2x 2 y 2
4
Solutions:
a. y, x
y 1 x 1 2y 2 x 2
x, y
x, ay bz
a x, y b x, z
2
2
x, x
x 1 2x 2 0 and 0 only if x 1 x 2 0
b. y, x
y 1 x 1 2y 1 x 2 2y 2 x 2
x, y
c. x, ay
a 2 x 21 y 21 2a 2 x 22 y 22 a 2 x, y
a x, y
d. y, x
y 1 x 1 y 1 x 2 y 2 x 1 2y 2 x 2
x, y
x, ay bz
a x, y b
2
2
2
2
x, x
x 1 2x 1 x 2 2x 2
x1 x2
x 2 0 and 0 only if x 1
e. y, x
y 1 x 1 y 1 x 2 y 2 x 1 2y 2 x 2
x, y
x, ay bz
a x, y b
x, x
x 21 2x 1 x 2 2x 22
x 1 x 2 2 x 22 0 and 0 only if x 1
f. x, x
x 21 2x 1 x 2 x 22
x 1 x 2 2 0 but can 0 if x 1 x 2 0
g. x, x
x 21 2x 1 x 2 x 22
x 1 x 2 2 0 but can 0 if x 1
x2 0
2
2
h. x, x
x 1 x 2 0 if x 2 x 1 0
10.
x, z
x2
x, z
x2
0
0
20 points
Let M 2, 3 be the vector space of 2 3 matrices.
Consider the subspace V Span A 1 , A 2 , A 3 , A 4 where
A1
1 1
0
2 3
2
, A2
2 2
0
4 6
4
1 0 2
, A3
, A4
3 0 0
1 2
6
5 6
4
Find a basis for V. What is the dim V ?
Solution: To see if A 1 , A 2 , A 3 , A 4 are linearly independent, we write:
aA 1 bA 2 cA 3 dA 4 0 and solve for a, b, c, d.
a
1 1
0
2 3
2
b
2 2
0
4 6
4
a
2b
c
d
a
2b
2d
2a
4b
3c
5d 3a
6b
6d
1
2
1
1
0
1
2
0
2
0
0
0
2
6
0
2
4
3
5
0
R4
3
6
0
6
0
2
4
0
4
0
c
2c
2a
1 0 2
d
3 0 0
6d
4b
4d
1 2
6
0 0 0
5 6
4
0 0 0
0 0 0
6 equations
0 0 0
4 unknowns
1 2
1
1
0
0 0
1
3
0
0 0
2
6
0
R3
2R 1
0 0
1
3
0
R5
3R 1
0 0
3
9
R6
2R 1
0 0
2
6
R2
R1
R1
R2
1 2 0 2
0
0 0 1 3
0
2R 2
0 0 0 0
0
R4
R2
0 0 0 0
0
0
R5
3R 2
0 0 0 0
0
0
R6
2R 2
0 0 0 0
0
a
2b 2d 0, c 3d 0
a
2r 2s b r c
3s d s
2r 2s A 1 rA 2 3sA 3 sA 4 0 So they are not linearly independent.
Further, with r 1 and s 0: A 2 2A 1 and with r 0 and s 1: A 4 2A 1 3A 3
So A 1 , A 2 also span and the above computation without columns 2 and 4 shows they are linearly
independent. So A 1 , A 2 is a basis and dim V 2.
5
11.
28 points
Compute
F dS over the hemisphere z
x2
25
y2
oriented upward, for the
H
vector field F
x 3 4y 2 4z 2 , 4x 2 y 3 4z 2 , 4x 2 4y 2
HINT: Use Gauss’ Theorem by following these steps:
z3 .
a. Write out Gauss’ Theorem for the Volume, V, which is the solid hemisphere
0 z
25 x 2 y 2 . Split up the boundary, V, into two pieces, the hemisphere, H, and
the disk, D, at the bottom. State orientations. Solve for the integral you want.
b. Compute the volume integral using spherical coordinates.
c. Compute the other surface integral over D by parametrizing the disk, computing the tangent
vectors and normal vector, checking the orientation, evaluating the vector field on the disk and
doing the integral.
d. Solve for the original integral.
Solution:
a. Gauss’ Theorem:
F dV
F dS
F dS
V
V
F dS
H
D
H is oriented upward. D is oriented downward.
F dS
F dV
H
F dS
V
D
2
b. In spherical coordinates: dV
2
/2
5
F dV
2
3
0
V
0
2
and
sin
sin d d d
er
e
F
r sin
x3
4y 2
2
sin
0
r cos
0
F dS
d.
H
er
y3
5
5
0
0
F dV
V
F dS
2
.
6 1 54
5
3750
0
r cos , r sin , 0 .
e
4z 2 , 4x 2
2
3
5
5
/2
cos
î0
4y 2
4r 3 dr d
z3
2
r4
r sin 2
k r cos 2
0
Reverse to point down:
F N dr d
0
D
N
4z 2 , 4x 2
F dS
3z 2
0
k
cos
3 2
3y 2
0
c. The disk may be parametrized by R r,
î
3x 2
F
N
0, 0, r
, , 4r 2
03
5
F N
0, 0, r
4r 3
1250
0
0
3750
1250
2500
D
6