Math 2270, Fall 2014
Instructor: Thomas Goller
8 September 2015
Test #1 Solutions
1
Solving a Linear System (10 points)
Consider the homogeneous linear system
x1
2x1
2x2 + x3 = 0
3x2 x3 = 0
(a) Write the linear system as a matrix equation A~x = ~b. (2 points)
Solution:

1
2
2
3
2 3

x
1 4 15
0
x2 =
1
0
x3
(b) Write the linear system as a vector equation x1~a1 + x2~a2 + x3~a3 = ~b. (2 points)
Solution:
x1

1
+ x2
2

2
+ x3
3


1
0
=
1
0
(c) Compute the reduced echelon form of the augmented matrix of the linear system. (3
points)
Solution:

1
2
2
3
1 0
1 0

1
!
2R1
0
2
1

1 0 +2R2
1 0
!
3 0
0 1
5 0
3 0
(d) Compute the solution set of the linear system and write the solution set as the span of
a vector. (3 points)
Solution:
8 2 39
3
2 3
5x3
5
< 5 =
~x = 43x3 5 = x3 435 , so the set of all solutions is Span 435 .
:
;
x3
1
1
2
Math 2270, Fall 2014
2
Instructor: Thomas Goller
8 September 2015
Inventions (10 points)
(a) Invent a linear system of two equations in two variables that has infinitely many solutions.
(2 points)
Possible solution:
x1 + x2 = 3
2x1 + 2x2 = 6
2 3
2 3
1
0
(b) Invent a vector ~v with no zero entries that is a linear combination of 415 and 415. (2
0
1
points)
Possible solution:
2 3
1
~v = 425
1
(c) Invent a 2 ⇥ 4 matrix A whose columns do not span R2 . (2 points)
Possible solution:

A=
2
1 1 1 1
1 1 1 1
3
1 0 
x
(d) Invent a vector ~b in R3 so that 40 15 1 = ~b has no solution. (2 points)
x2
0 0
Possible solution:
2 3
0
~b = 405
1
Comment: Any ~b with third entry nonzero will do.
2 3

1
1
4
5
(e) Invent a matrix A so that 2 is a solution of A~x =
. (2 points)
4
3
Possible solution:
A=

1 0 0
4 0 0
Math 2270, Fall 2014
3
Instructor: Thomas Goller
8 September 2015
Solving Another Linear System (10 points)
2
0
6 6
Consider the linear system 6
41
1
3
2
3
0 
0
6
7
127
7 x1 = 6 127.
4 2 5
2 5 x2
2
2
(a) Compute the reduced echelon form of the augmented matrix of the linear system. (3
points)
Solution:
2
0
6 6
6
41
1
0
12
2
2
3
2
0
6
127
7!6
5
4
2
2
1
6
1
0
3
2
2
1 2
6
127
+6R
0 0
1
7
!6
5
4
2
R1
0 0
0
0 0
2
12
2
0
3
2
07
7
05
0
(b) Write the solution set of the system in parametric vector form. (3 points)
Solution:
~x =

2
2x2
x2
= x2


2
2
+
1
0
(c) Sketch the solution set in the plane R2 . (2 points)

2
Description of solution: Your sketch should be the line containing
in the direction
0



2
2
0
, which is the line containing
and
. You should label two points on the line
1
0
1
and label the axes x1 and x2 .
(d) Is this solution set the span of a vector? Why or why not? (2 points)
Solution: No. The solution set is a line that does not contain the origin

0
.
0
Comment: The span of a vector always contains the origin because you can choose the
scalar to be 0.
Math 2270, Fall 2014
4
Instructor: Thomas Goller
8 September 2015
Span (10 points)
(a) Sketch the set Span
⇢
1
3
in the plane R2 . (2 points)

0
Description of solution: Your sketch should be the line containing the origin
and
0

1
. You should label two points on the line and label the axes (x, y or x1 , x2 ).
3
(b) Sketch the set Span
⇢

1
0
,
3
1
in R2 . (2 points)
Description of solution: Your sketch should be the entire plane R2 , which you can depict
by shading. You should label the axes (x, y or x1 , x2 ).
(c) What can you deduce about vectors ~v and w
~ in R3 if Span{~v , w}
~ is just one point? (2
points)
Solution:
2 3
0
4
~v = 05
0
2 3
0
4
and w
~ = 05
0
Comment: If ~v or w
~ are nonzero, then they will span at least a line!
(d) What can you deduce about vectors ~v and w
~ in R3 if Span{~v , w}
~ is a line? (2 points)
Solution: ~v is a scalar multiple of w
~ or w
~ is a scalar multiple of ~v .
Comment: You need both cases because either vector could be ~0.
(e) Suppose you know that Span{~v , w}
~ is a plane in R3 . Suppose ~u in R3 is not on that
plane. What is Span{~v , w,
~ ~u}? (2 points)
Solution: The entire space R3 .
Comment: Since ~u is not in Span{~v , w},
~ including ~u in the span enlarges the set by
one dimension (we will make this precise later in the course). In other words, adding
multiples of ~u allows the linear combinations to move o↵ of the plane and produce any
point of R3 .
Math 2270, Fall 2014
5
Instructor: Thomas Goller
8 September 2015
Echelon Form (10 points)
2
0 1 4
4
Consider the matrix A = 0 0 1
0 0 0
0
2
0
3
3
35.
0
(a) Describe a row operation in words that would put A into reduced echelon form. (2
points)
Solution: Subtract 4 times row 2 from row 1.
(b) Which columns of A are the pivot columns? (2 point)
Solution: Columns 2 and 3.
(c) In the equation A~x = ~b, which are the free variables? (2 point)
Solution: x1 , x4 , x5 .
(d) What can you say about ~b in R3 if you know that A~x = ~b has no solution? (1 points)
Solution: The third entry of ~b is nonzero.
(e) Describe the set of all ~b in R3 for which A~x = ~b has exactly one solution. (1 points)
Solution: The empty set {}.
Comment: A~x = ~b never has a unique solution when there are free variables!
(f) Suppose that a matrix B can be row reduced to get the echelon matrix A. What can
you deduce about the first column of B? What can you deduce about the last row of
B? (2 points)
2 3
0
Solution: The first column of B must be 405. The last row of B is a linear combination
0
of the first two rows of A.
Comment: You can’t say that the last row of B is a multiple of another row of B or
even that the last row of B is a linear combination of the first two rows of B because
row reduction from B to A could have involved a row swap.