solutions

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
3a1 3a2 3a3
5b1 5b2 5b3
7c1 7c2 7c3
k det
a1 a2 a3
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.
3a1 3a2 3a3
a1 a2 a3
det 5b1 5b2 5b3 3det 5b1 5b2 5b3 15det
105det
7c1 7c2 7c3
a1 a2 a3
b1 b2 b3
c1 c2 c3
7c1 7c2 7c3
. Therefore, k 105.
a1 a2 a3
b1 b2 b3
7c1 7c2 7c3
———————————————————————-
2. Let A,B,C be 3 3 matrices. Suppose detA 7,detB 3, and detC 0. Calculate
the following. (Be sure to explain why.)
———————————————————————a.
detAB detAdetB by Theorem 6 on page 196. Therefore, detAB 21
b. detC T detC by Theorem 5 onpage 196. Therefore, detC T 0.
c. detA 1 1
det A by exercise 31 on page 200. Therefore, detA1 1
7 .
———————————————————————-
3. Let W be the set of all vectors of the form
4a 2b
3c
5b 2c
2a 3b c
where a,b, andcrepresent
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.
———————————————————————-

The set W can be written as
4a 2b
3c
5b 2c
2a 3b c
4 2 0
Let S
0
0
,
0
5
,
3
2
. Since a,b, andcrepresent arbitrary real
2 3 1
numbers, W is a set of all linear combinations the vectors in S or the span of the vectors in S.
a
4
0
0
2
b
2
0
5
3
———————————————————————-
4. Prove that the set of all vectors of the form a
1
1
0
b
subspace of 3 . Use the definition of subspace on page 220.
i) 0 0
1
1
0
ii) Let u a1
0
1
0
———————————————————————-
0
1
1
0
0
1
0
b1
so the zero vector is in the given set.
0
1
0
and v a2
1 0
1 0
a1 1 b1 1 a2 1 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 0
2
1
1
0
b2
0
1
0
c
0
3
2
1
,a,b , forms a
a1 a2
. Then u v
1
1
0
.
b1

1 0
1 0
ku k a 1 b 1 ak 1 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 4
0
5. Let A
and w .
1 7
9
———————————————————————-
Is w in NulA?
Aw
NulA.
2 4
1 7
Is w in ColA?
0
9
36
63
. Since the product is not the zero vector, w is not in
Solving the equation Ax b or x we write the augmented matrix:
1 0 2
0 1 1
. Since the system is consistent w is in ColA.
———————————————————————-
2 4 0
1 7 9
6. Assume that A is row equivalent to B. Find bases for NulA and ColA.
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
———————————————————————-
AbasisforNulA :
3

x2
1 3 0 4 1
0 0 1 3 4
0 0 0 0 0
3
1
0
0
0
x4
4
0
3
1
0
x5
Therefore the nullspace basis is
x1
x2
x3
x4
x5
1
0
4
0
1
.
3x2 4x4 x5
x2
3x4 4x5
x4
x5
3 4
1 0
0 , 3 ,
0 1
0 0
AbasisofColA :
Since there are pivots in columns 1 and 3, we choose columns 1 and 3 from A to form the
basis of ColA.
5 2
That is ColA 2 , 0
1 8
———————————————————————-
7. The set B 1,1 t,t t2 ,t t3is a basis for P3 . Find the coordinate vector of
pt 2t t2 8t3 .
———————————————————————-
The change of coordinate matrix, PB
x B ,weusetheaugmentedmatrix
4
1
0
4
0
1
1 1 0 0
0 1 1 1
0 0 1 0
0 0 0 1
1 1 0 0 0
0 1 1 1 2
0 0 1 0 1
0 0 0 1 8
. To find the coordinate vector
1 0 0 0 5
0 1 0 0 5
0 0 1 0 1
0 0 0 1 8
.

Therefore, x B
5
5
1
8
———————————————————————-
8.
———————————————————————a.
Suppose a 5 8matrixAhas 5 pivot columns.
What is dimNulA? There are 3 free variables and therefore dimNulA 3.
What is rankA? Since there are 5 pivot columns rankA 5.
b. The null space of a 20 18 matrix A is 10-dimensional.
What is the dimension of ColA?
Since there are 18 columns, i.e., n 18, rankA n dimNulA 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 NulA?
Since the greatest dimension of the column space is 50, and
rankA dimNulA n 50, the smallest possible dimension of NulA 0.
d. A 38 50 matrix A has rank 30.
Find dimNulA.
By the Rank Theorem, rankA dimNulA n
30 dimNulA 50 dimNulA 20.
Find dimRowA.
By the Rank Theorem, the dimension of the column space equals the dimension of the row
space. This implies RowA 30.
Find RankA T .
RankA T dimColA T dimRowA 30.
5
Explain all.