You must show all your work. The number of points earned on
each problem will be determined by how well you have justified your
work.
1. Find the value of k that satisfies the following equation.
———————————————————————det
3a 1
3a 2
3a 3
5b 1
5b 2
5b 3
a1 a2 a3
 k  det
7c 1 7c 2 7c 3
b1 b2 b3
c1 c2 c3
———————————————————————By Theorem 3 on page 192, if one row of a matrix is multiplied by a scalar, then the
determinant is also.
3a 1 3a 2 3a 3
a1
a2
a3
a1
a2
det 5b 1 5b 2 5b 3
 3 det 5b 1 5b 2 5b 3
 15 det b 1
b2
7c 1 7c 2 7c 3
7c 1 7c 2 7c 3
a3
b3
7c 1 7c 2 7c 3
a1 a2 a3
105 det
b1 b2 b3
. Therefore, k  105.
c1 c2 c3
———————————————————————2. Let A, B, C be 3  3 matrices. Suppose det A  7, det B  3, and det C  0. Calculate
the following. (Be sure to explain why.)
———————————————————————a. detAB  det Adet B by Theorem 6 on page 196. Therefore, detAB  21
b. detC T   detC by Theorem 5 onpage 196. Therefore, detC T   0.
c. detA 1  
1
det A
by exercise 31 on page 200. Therefore, detA 1   17 .
———————————————————————4a  2b
3. Let W be the set of all vectors of the form
3c
5b  2c
where a, b, and c represent
2a  3b  c
arbitrary real numbers. Either find a set S of vectors that spans W or give an example to show
that W is not a vector space.
———————————————————————-
4a  2b
The set W can be written as
3c
0
a
5b  2c
2
0
Let S 
0
c
5
2
0
3
3
2
.
1
0
0
,
0
b
0
2a  3b  c
4
2
4
,
5
3
. Since a, b, and c represent arbitrary real
2
2
3
1
numbers, W is a set of all linear combinations the vectors in S or the span of the vectors in S.
———————————————————————-
1
4. Prove that the set of all vectors of the form a
0
b
1
0
subspace of  3 . Use the definition of subspace on page 220.
, a, b  , forms a
1
0
———————————————————————1
i) 0  0
0
0
1
1
0
0
1
ii) Let u  a 1
0
 b1
1
0
1
a1
1
so the zero vector is in the given set.
1
and v  a 2
1
0
1

1
 b2
1
0
0
 b1
0
a2
1
2
0
. Then u  v 
0
0
 b2
1
0
0
0
0
which has the required form. So, u  v is in the required set.
1
0
iii) u  a 1
 b 1 . Then
0
1
1
 a 1  a 2 
1
0
 b 1 
1
ku  k a
0
b
1
1
 ak
1
1
0
 bk
1
0
0
0
0
which again has the required form. So, ku is in the required set.
By the definition on page 220 the set is a subset of  3 .
———————————————————————2
5. Let A 
4
and w 
0
.
1 7
9
———————————————————————Is w in NulA?
2
Aw 
4
1 7
0
9

36
63
. Since the product is not the zero vector, w is not in
NulA.
Is w in ColA?
Solving the equation Ax  b or x we write the augmented matrix:
1 0
2
0 1 1
2
4 0
1 7 9
. Since the system is consistent w is in ColA.
———————————————————————-
6. Assume that A is row equivalent to B. Find bases for NulA and ColA.
5 15 2 26 3
1 3 0 4 1
A
2 6 0 8 2 , B 
0 0 1 3 4
1 3
8 28 31
0 0 0 0 0
———————————————————————A basis for NulA :
3

1 3 0 4 1
x2
0
0
1 3
4
0
0
0 0
0

x1
3x 2  4x 4  x 5
x2
x2

x3
x4
x4
x5
x5
3
4
1
1
0
0
0
 x4
3
 x5
1
0
0
0
1
Therefore the nullspace basis is
.
4
0

3x 4  4x 5
3
4
1
1
0
0
0 , 3 , 4
0
1
0
0
0
1
A basis of ColA :
Since there are pivots in columns 1 and 3, we choose columns 1 and 3 from A to form the
basis of ColA.
5
2
That is ColA 
2 ,
0
1
8
———————————————————————7. The set B  1, 1  t, t  t 2 , t  t 3  is a basis for P 3 . Find the coordinate vector of
pt  2t  t 2  8t 3 .
———————————————————————1 1 0 0
The change of coordinate matrix, P B 
0 1 1 1
. To find the coordinate vector
0 0 1 0
0 0 0 1
1 1 0 0 0
x B , we use the augmented matrix
0 1 1 1 2
0 0 1 0 1
0 0 0 1 8
4
1 0 0 0

5
0 1 0 0 5
0 0 1 0 1
0 0 0 1
8
.
5
Therefore, x B 
5
1
8
———————————————————————-
8.
———————————————————————a. Suppose a 5  8 matrix A has 5 pivot columns.
What is dimNulA? There are 3 free variables and therefore dimNulA  3.
What is rankA? Since there are 5 pivot columns rankA  5.
b. The null space of a 20  18 matrix A is 10-dimensional.
What is the dimension of ColA?
Since there are 18 columns, i.e., n  18 , rankA  n  dim NulA  18  10  8.
c. If A is a 200  50 matrix, what is the largest possible dimension of the row space?
By Theorem 14, the Rank Theorem, the dimension of the column space equals the
dimension of the row space. There are at most 50 pivots, so the largest dimension of the
column and row spaces is 50.
What is the smallest possible dimension of NulA?
Since the greatest dimension of the column space is 50, and
rankA  dim NulA  n  50, the smallest possible dimension of NulA  0.
d. A 38  50 matrix A has rank 30.
Find dimNulA.
By the Rank Theorem, rankA  dim NulA  n 
30  dim NulA  50  dim NulA  20.
Find dimRowA.
By the Rank Theorem, the dimension of the column space equals the dimension of the row
space. This implies RowA  30.
Find RankA T .
RankA T   dim ColA T   dim RowA  30.
Explain all.
5