Math 2270, Fall 2014
Instructor: Thomas Goller
16 December 2014
Final Exam Solutions
1
The Four Subspaces (10 points)
Let A =

2
2
0 1
.
1 0
(a) Compute the rank r of A. (1 point)
Solution:

2
2

0 1
2
!
1 0
0
0 1
,
1 1
has 2 pivots, so r = 2.
(b) Use r to compute the dimensions of the four fundamental subspaces N (A), C(A), C(AT ),
N (AT ). (2 points)
Solution: dim N (A) = 1, dim C(AT ) = 2, dim C(A) = 2, dim N (AT ) = 0.
(c) Which pairs of subspaces are orthogonal? (1 point)
Solution: N (A) and C(AT ); C(A) and N (AT ).
(d) Compute bases for the four fundamental subspaces of A. (6 points)
8 2 39
8 2 3 2 39
0 =
< 1 =
< 2
T
4
5
4
5
4
2
0 ,
15 .
Solution: Basis for N (A):
. Basis for C(A ):
:
;
:
;
2
1
1
⇢

2
0
Basis for C(A):
,
. Basis for N (AT ): {}
2
1
Math 2270, Fall 2014
2
Instructor: Thomas Goller
16 December 2014
Inverse (10 points)
2
1
4
Let A = 3
0
0
2
4
3
2
0 5.
10
⇥
⇤
(a) Compute the inverse A 1 of A using Gauss-Jordan elimination on A I or the cofactor
formula det1 A C T . (8 points)
Solution
2
1
43
0
(Gauss-Jordan elimination):
3
2
0
2 1 0 0
1 0
2 1
5
4
2 0 0 1 0 ! 0
2 6 3
4
10 0 0 1
0
4
10 0
2
3
2
1 0 0
5
2
1
1
4
5
4
2 0
15
5
3 ! 0
! 0
0
0 2 6
2
1
0
3
2
0 0
1
5
4
1 0 ! 0
0 1
0
3
0 2 1 0 0
2 6 3 1 05
0 2 6 2 1
3
0 0
5
2
1
1 0 15/2 5/2 3/25
0 1
3
1 1/2
(b) Check your inverse is correct by showing A 1 A = I. (2 points)
Solution:
2
32
5
2
1
1
415/2 5/2 3/25 4 3
3
1 1/2
0
0
2
4
3 2
3
2
1 0 0
0 5 = 4 0 1 05
10
0 0 1
Math 2270, Fall 2014
3
Instructor: Thomas Goller
16 December 2014
Subspaces (10 points)
Recall that a subset S of a vector space is called a subspace if two conditions hold:
(i) For every vector ~v in S, every c~v is still in S.
(ii) For every two vectors ~v and w
~ in S, the sum ~v + w
~ is still in S.
For each S defined below, state whether condition (i) holds, whether condition (ii) holds,
and whether S is a subspace. (You do not need to show any other work.)
(a) The line S in R2 with equation 2x + 4y = 0. (2 points)
Solution: (i) holds, (ii) holds, S is a subspace.

x
(b) The set S of all solutions
in R2 to the equation
y

1
2
7 14


x
2
=
. (2 points)
y
14
Solution: (i) does not hold, (ii) does not hold, S is not a subspace.
(c) The set S of all vectors

x
in R2 such that x and y are integers. (2 points)
y
Solution: (i) does not hold, (ii) holds, S is not a subspace.
(d) The union S of the coordinate axes in R (the set of all vectors
2
(2 points)

x
with x = 0 or y = 0).
y
Solution: (i) holds, (ii) does not hold, S is not a subspace.
(e) The set S = span(~v1 , . . . , ~vn ), where ~v1 , . . . , ~vn are vectors in a vector space V . (2
points)
Solution: (i) holds, (ii) holds, S is a subspace.
Math 2270, Fall 2014
4
Instructor: Thomas Goller
16 December 2014
Inventions (10 points)
(a) Invent two vectors ~v1 , ~v2 in R3 so that span(~v1 , ~v2 ) is a line. (2 points)
2 3
2 3
1
1
4
5
4
25.
Possible solution: ~v1 = 2 , ~v2 =
3
3
T
(b) Invent two vectors ~v and w
~ so that ~v w
~ =
Possible solution: ~v =

2
1
4
2
6
. (2 points)
3
2 3
1
2
,w
~ = 425.
1
3

(c) Invent a system of two linear equations in x and y that has no solution. (2 points)
Possible solution: x + y = 0, x + y = 3.
(d) Invent a 2 ⇥ 2 matrix A such that N (A) = C(A). (2 points)
Possible solution: A =

0 1
.
0 0
(e) Invent a matrix A so that det(A) =
2
3
1 0 0
Possible solution: A = 4 0 1 05.
0 0 1
1 and det(3A) =
27. (2 points)
Math 2270, Fall 2014
5
Instructor: Thomas Goller
16 December 2014
Closest Line to Three Points (10 points)
Let b = C + Dt be the equation for a line L in R2 .
(a) Write down the three
 linear
 equations

in C and D that would have to hold for L to pass
0
1
2
t
through the points
=
,
,
. (2 points)
b
2
0
0
Solution:
2 = C, 0 = C + D, 0 = C + 2D.
(b) Convert those linear equations into a matrix equation A

C
= ~b. (1 point)
D
2
3
2 3
1 0 
2
C
4
5
4
Solution: 1 1
= 0 5.
D
1 2
0
T
(c) Write down the new matrix equation A A
Solution:


C
= AT~b. (2 points)
D
2
3
2 3



1 0
2
1 1 1 4
3
3
1
1
1
2
4 0 5=
1 15 =
;
0 1 2
3 5
0 1 2
0
1 2
0



3 3 C
2
=
3 5 D
0
(d) Solve the new matrix equation. (3 points)
Solution:

C
=
D
1
6

5
3
3
3

2
=
0
1
6

10
=
6

5/3
1
(e) Draw a graph containing the three points and the closest line L. (2 points)
Description of solution: Plot the three points and the line L with equation b =
5/3 + t.
Math 2270, Fall 2014
6
Instructor: Thomas Goller
16 December 2014
Projection (10 points)
2
3
2 3
1
1
3
~
4
5
4
Let S be the plane in R spanned by vectors ~a = 0 and b = 25.
1
3
~ B
~ for S. (3 points)
(a) Compute an orthogonal basis A,
Possible solution:
2
3
1
~ = 4 0 5;
A
1
2 3
1
~
4
B = 25
3
2
3 2 3
1
2
24 5 4 5
0 = 2
2
1
2
~ and B
~ to get an orthonormal basis ~q1 , ~q2 for S. (1 point)
(b) Normalize your vectors A
Solution:
2
3
1
~q1 = p12 4 0 5 ,
1
2 3
1
~q2 = p13 415 .
1
(c) Compute the matrix P = QQT that projects vectors orthogonally onto S . (4 points)
Solution:
p
p 3
1/ 2 1/p3  p
0p
4 0
5 1/p2
p 1/p3 1/ 3 1/ 3
1/ 2 1/ 3
2
2
p
5/6 1/3
1/p 2
4
= 1/3 1/3
1/ 3
1/6 1/3
3
1/6
1/3 5
5/6
2
3
1
(d) Show that 4 0 5 is an eigenvector of P . What is the eigenvalue ? (2 points)
1
Solution:
2
5/6 1/3
4 1/3 1/3
1/6 1/3
32 3 2
3 2 3
1/6
1
5/6 + 1/6
1
1/3 5 4 0 5 = 4 1/3 1/3 5 = 4 0 5 ,
5/6
1
1/6 5/6
1
= 1.
Math 2270, Fall 2014
7
Instructor: Thomas Goller
16 December 2014
Di↵erential Equations (10 points)
Consider the following population model. Let b(t) denote the population of bananas and
g(t) the population of gorillas at time t. The growth rates of the two populations are
db
= 4b
dt
10g
and
dg
= 15 b + g.
dt
(a) Write these growth rates as a di↵erential equation of the form
(b) Compute the eigenvalues
1,
2
d~
u
dt
= A~u. (2 points)
and corresponding eigenvectors ~x1 , ~x2 of A. (5 points)
(c) Initially, there are b(0) = 60 bananas and g(0) = 10 gorillas. Compute the scalars C1 , C2
that give the unique solution ~u(t) = C1 e 1 t~x1 + C2 e 2 t~x2 . (3 points)
Math 2270, Fall 2014
8
Instructor: Thomas Goller
16 December 2014
Diagonalization (10 points)
Let A =

1 4
.
2 5
(a) Compute the eigenvalues
1,
of A. (2 points)
2
Solution:
1
4
2
2
=
5
4
5+8=(
3)(
1),
1
= 3,
2
= 1.
(b) Compute independent eigenvectors ~x1 , ~x2 of A. (3 points)
Solution:
A
3I =
A
I=


4 4
!
2 2
2 4
!
2 4
(c) Write down the factorization A = S⇤S
the eigenvalue matrix. (2 points)
Solution:
A=

1


1 1
,
0 0
1 2
,
0 0
~x1 =
~x2 =


1
1
2
1
, where S is a matrix of eigenvectors and ⇤ is



1 4
1 2 3 0
1
=
2 5
1 1 0 1
1
2
1
(d) Use your answer in (c) to compute A4 . Simplify completely. (3 points)
Solution:

1 2
A =
1 1
4

81 0
0 1

1
1


2
81 2
1
=
1
81 1
1
2
=
1

79 160
80 161
Math 2270, Fall 2014
9
Instructor: Thomas Goller
16 December 2014
Linear Transformations in R2 (10 points)
Suppose you know R
2
T
/
R is linear and T
2
✓
2
1
◆
=

7
and T
3
✓
1
1
◆
=

6
.
1
(a) Compute each of the following if you can, or state that not enough information is given:
(3 points)
Solution:
✓ ◆
✓  ◆
✓ ◆


0
2
2
7
0
(i) T
=T 0
= 0T
=0
=
0
1
1
3
0
(ii) T
✓
1
0
◆
(iii) T
✓
0
1
◆
=T
✓
=T
✓ 
2
1

1
1
◆
=T
✓
 ◆
2
1
+2
=
1
1
2
1

◆
T
✓
1
1
◆
=


7
3
6
=
1

1
4


7
6
5
+2
=
3
1
5
(b) T acts as multiplication by a matrix A. Use your answer in (a) to find A. (2 points)
Solution: A =

1 5
4 5




1
0
1
0
(c) Draw the square with vertices
,
,
,
and draw the parallelogram that you
0
0
1
1
get when T transforms that square. (4 points)

0
Description of solution: The parallelogram should have vertices at
,
0

6
.
1


1
5
,
, and
4
5
(d) Compute det A. This is the area of the parallelogram you just drew! (1 point)
Solution: det A = 5 + 20 = 25
Math 2270, Fall 2014
10
Instructor: Thomas Goller
16 December 2014
More Inventions (10 points)
(a) Invent a 2⇥2 matrix A that has an eigenvalue of multiplicity 2 but only one independent
eigenvector. (2 points)
Possible solution: A =

1 9
.
0 1
(b) Invent a matrix A such that no other matrix is similar to A. (2 points)
Possible solution: A =

1 0
.
0 1
(c) Invent a basis B = {~v1 , ~v2 } for R such that the B-coordinates of the vector

1
. (2 points)
1
2
Possible solution: ~v1 =
p
2
are
3
p

0
2
, ~v2 =
.
3
0
⇢

1
2
(d) Let B =
,
. Invent a matrix M that changes the coordinates of vectors from
2
3
standard coordinates to B-coordinates. (2 points)
Possible solution: M =
1
7

3
2
2
1
(e) Let V be the vector space of all polynomials in x of degree
2 3 2. Invent a basis B =
1
{~v1 , ~v2 , ~v3 } of V such that the B-coordinates of x + x2 are 415. (2 points)
1
Possible solution: ~v1 = 1, ~v2 =
1 + x, ~v3 = x2 .