MATH 311
Exam 1
Section 502
Fall 2000
Solutions P. Yasskin
"1 0 3
1. (10 points) Find the inverse of A "1 0 3
1 0 0
0
1 4
0 1 0
1
1 0
0 0 1
1
1 0
0
0
1 4
1
1 0
.
R3
1 0 "4
0 "1 1
0 1
0
4
1
0
0 0 "1
1 "1 1
0 0 1
1 0 "4
0
"1 1
R1 4R3
1 4
0 1 0
0 1
4
0
1
R2 " 4R3
"1 0 3
1 0 0
0 0
1
"1 1 "1
1 1 0
0 0 1
0 1 4
0 1 0
0 1 3
1 0 1
R1
R3 R1
R1 " R2
1 0 0
0 1 0
R3 " R2
0 0 1
A "1 "R3
0
"4 3 "3
4 "3 4
"1 1 "1
"4 3 "3
4 "3 4
"1 1 "1
0 0 2
Use it to solve XA 2 1 0
.
0 2 0
0 0 2
XA B
®
X
BA "1
2 1 0
0 2 0
"4 3 "3
4 "3 4
"1 1 "1
"2 2 "2
"4 3 "2
8 "6 8
1
2. (10 points) Consider the polynomials
p 1 Ÿx 1 " x 2
p 2 Ÿx 2 " x " x 2
p 3 Ÿx 1 " x
and the vector space
W SpanŸp 1 , p 2 , p 3 .
Find a subset of £p 1 , p 2 , p 3 ¤ which is a basis for W. Prove it spans W and is linearly independent.
ap 1 bp 2 cp 3 0
"a"b 0
®
"1 "1 0
0 "1 "1
0
1
0
2
1
0
aŸ1 " x 2 bŸ2 " x " x 2 cŸ1 " x 0
®
"R1
"R2
R3 R1
"b"c 0
1 1 0
0
0 1 1
0
0 1 1
0
R1 " R2
R3 " R2
a 2b c 0
1 0 "1
0
0 1
1
0
0 0
0
0
R1 " R2
at
b "t
R3 " R2
ct
Let t 1. Then
p1 " p2 p3 0
or
p 3 "p 1 p 2
Claim the basis is £p 1 , p 2 ¤. They span because
ap 1 bp 2 cp 3 ap 1 bp 2 cŸ"p 1 p 2 Ÿa " c p 1 Ÿb c p 2
They are linearly independent because
ap 1 bp 2 0
®
®
"a"b 0
®
aŸ1 " x 2 bŸ2 " x " x 2 0
"b 0
a 2b 0
b0 & a0
3. Consider the vector space P 3 , the set of polynomials of degree 3 or less?
S (5 points)
a. A £
b. B £
c. C £
d. D £
e. E £
Scantron #1 Which of the following is NOT a subspace of P 3 ?
p P 3 | pŸ 0 0 ¤
p P 3 | pŸ 1 0 ¤
p P 3 | pŸ0 pŸ1 ¤
p P 3 | pŸ0 pŸ1 0 ¤
Correct
p P 3 | pŸ 0 1 ¤
E is not a subspace because if p, q E then pŸ0 1 and qŸ0 1. So Ÿp q Ÿ0 2 and p q E.
2
4. Consider the vector space R of all positive real numbers with the operations of
Vector Addition:
x ¥ y xy
(real number addition)
)
Scalar Multiplication:
)(x x
(real number exponentiation)
S (5 points) Scantron #2 Translate the vector identity
0(x 0
into ordinary arithmetic.
a. 1 x 1
Correct
b. x 0 1
x
c. 0 0
d. x 1 x
e. 0 x 1
0 ( x x0
and
0 1.
So x 0 1.
1 "1 0
5. Consider the linear map L : R 3 ¯ R 4 given by LŸ
x Ax where A 0
1
2
2
0
4
.
3 "1 4
2
S (10 points) Solve LŸx "1
2
.
4
1 "1 0
2
1 "1 0
2
0
1
2
"1
0
1
2
2
0
4
2
R3 " 2R1
0
2
4
3 "1 4
4
R4 " 3R1
0
2
4
"1
"2
"2
x 2z 1
®
y 2z "1
R1 R2
1 0 2
1
0 1 2
"1
R3 " 2R2
0 0 0
0
R4 " 2R2
0 0 0
0
x 1 " 2t
®
y "1 " 2t
zt
S (5 points) Scantron #3 Describe the solution set:
a.
b.
c.
d.
e.
No Solutions
Unique Solution (Point in R 3 )
.-Many Solutions (Line in R 3 )
.-Many Solutions (Plane in R 3 )
.-Many Solutions (All of R 3 )
Correct
There is one parameter.
S (5 points) Scantron #4 Is L a one-to-one function?
a. Yes
b. No
Correct
b.
There is more than one solution to LŸx 3
1 "1 0
6. Again consider the linear map L : R 3 ¯ R 4 given by LŸ
x Ax where A 0
1
2
2
0
4
.
3 "1 4
1
S (10 points) Solve LŸx 1
1
.
1
®
1 "1 0
1
1 "1 0
1
0
1
2
1
0
1
2
1
2
0
4
1
R3 " 2R1
0
2
4
3 "1 4
1
R4 " 3R1
0
2
4
"1
"2
R1 R2
1 0 2
1
0 1 2
"1
"3
"4
R3 " 2R2
0 0 0
R4 " 2R2
0 0 0
No Solution
S (5 points) Scantron #5 Describe the solution set:
a.
b.
c.
d.
e.
No Solutions
Correct
Unique Solution (Point in R 3 )
.-Many Solutions (Line in R 3 )
.-Many Solutions (Plane in R 3 )
.-Many Solutions (All of R 3 )
S (5 points) Scantron #6 Is L an onto function?
a. Yes
1
b. No
Correct
1
1
is not in the image.
1
4
1 "1 0
7. Again consider the linear map L : R 3 ¯ R 4 given by LŸ
x Ax where A 0
1
2
2
0
4
.
3 "1 4
S (5 points) Find KerŸL , the kernel (or null space) of L.
®
1 "1 0
0
1 "1 0
0
0
1
2
0
0
1
2
0
2
0
4
0
R3 " 2R1
0
2
4
0
3 "1 4
0
R4 " 3R1
0
2
4
0
x 2z 0
y 2z 0
R1 R2
1 0 2
0
0 1 2
0
R3 " 2R2
0 0 0
0
R4 " 2R2
0 0 0
0
x "2t
®
y "2t
®
"2t
"2t
KerŸL zt
S (5 points) Give a basis for KerŸL . (No proof)
t
"2
"2
1
S (5 points) What is the dimension of KerŸL ? (No proof)
dim KerŸL 1
5
1 "1 0
8. Again consider the linear map L : R 3 ¯ R 4 given by LŸ
x Ax where A 0
1
2
2
0
4
.
3 "1 4
S (5 points) Find ImŸL , the image (or range) of L.
1 "1 0
LŸx Ax 0
1
2
2
0
4
x"y
x
y 2z
y
3 "1 4
"1
0
2
1
,
"1
3
0
0
z
"1
3
2
4
4
0
,
0
1
y
2
3x " y 4z
1
ImŸL £LŸx ¤ Span
0
x
2x 4z
z
"1
1
2
4
4
S (5 points) Give a basis for ImŸL . (No proof)
The 3 vectors span. Are they independent?
x
1
"1
0
0
1 "1 0
0
0
1
2
0
0
1
2
0
2
0
4
0
3 "1 4
0
2
y
z
0
"1
3
4
4
0
®
0
x "2t
®
y "2t
zt
as in Problem 7. So they are not independent. Throw out the third vector and the first two are
independent.
"1
1
Basis is
0
2
3
,
1
0
"1
S (5 points) What is the dimension of ImŸL ? (No proof)
dim ImŸL 2
6