Group Solve 2 Solutions
Math 220
1. Consider a homogeneous system of seven linear equations in nine variables. Suppose
that there are two solutions of the system that are not multiples of one another, and all other
solutions are linear combinations of these solutions. Will the associated non-homogeneous
system have a solution for every possible choice of constants on the right sides of the
equations? Justify your answer. Mention any theorems that you use.
Let A be the 7 x 9 coefficient matrix for the system. By the definition of Nul A, the set of
solutions to the equation Ax = 0 is Nul A. Since this set contains two solutions of the system
that are not multiples of one another, the basis of Nul A consists of two linearly independent
vectors (see the box on the bottom of page 67). Thus, dim Nul A = 2.
By Theorem 14 Ch. 4, rank A + dim Nul A = n, so we have rank A + 2 = 9 and rank A = 7.
So, there are seven pivot positions in A. Thus, by Theorem 4 Ch 1, there is at least one
solution to Ax = b for all b in R7. Since the choices of constants on the right sides of the
equations are the entries of b, there is a solution to every associated non-homogeneous
system.
4
3
2 Suppose   b1,b2  where b1    , b 2    , and
5
6 
E  e1,e2  where e1  1 , e 2  0 .
0 
1
Find the change of basis coordinates matrix to go from β to the standard basis and from the
standard basis to β. Label your answers clearly. No explanations are needed if work is
clearly shown.
The matrix [b1 b2] changes vectors in β –coordinate system to vectors in the standard
3 4
coordinates. To go from β to the standard basis, use 
 . To go the other way, use
5 6
1
2 
3 4
 3
5 6  2.5  1.5 .




3. Assume A  a1 a2 a3 a4 a5  and B  b1 b2 b3 b4 b5  are row equivalent, where
2 2
0
7
 1
1
 2  3

0
1  1  5

A
and B  
 3  4
0
0  2  3



6 6
5
1
 3
0
a. Find a basis for the column space of A.
 1   2   0  
      
   2   3   1  
,
,






  2 

3

4

 3   6   5  
0
4 0  3
1 3 0
5
0
0 1  4

0
0 0
0
No explanations are needed.
b. Find a basis for Row A. No explanations are needed.
 1   0   0  
      
 0   1   0  


 4 ,  3,  0   or 1,0,4,0,3, 0,1,3,0,5, 0,0,0,1,4
 0   0   1  
      
 3  5   4 
c. Fill in the blank. Nul A is a 2 dimensional subspace of R5. No explanations are needed.
4. Determine whether the following sets are subspaces. Justify your answer. Mention any
theorems that you use.
 a  3b 




 a  b 

b. H  
: a, b  R 
  2a  b 

 4a 

a. The set is not a subspace since it is not closed under scalar multiplication. For example,
1
 1


u  1 is in the set, but  1u   1 is not.
1
 1
a 

 

a. G  b  : a  0, b  0, c  0
 c 

 

 a  3b 
  1   3 

 1   3  

     

    

 a  b 
  1  1

 1 1 
b. H  
: a, b  R   a
b
: a, b  R   Span ,   
  2a  b 
   2  1 

  2  1  






 4a 
  4  0 

4  0  
So, by Theorem 1 Ch 4, H is a subspace of R4.
5. Prove the Theorem: If A is an m x n matrix, then Nul A is a subspace of Rn.
Nul A is a subset of Rn, so we need to show that the following properties hold.
a. 0 is in Nul A,
b. Nul A is closed under vector addition,
c. Nul A is closed under scalar multiplication.
By definition, v  NulA iff Av = 0.
a. A0 = 0, so 0 is in NulA
b. Let u, v be in Nul A. Then Au = 0 & Av = 0.
A(u + v) = (by Th. 5 Ch 1) Au + Av = 0 + 0 = 0
So, Nul A is closed under vector addition.
c. Let u be in Nul A and c a scalar.
Then Au = 0.
Acu = (By Th. 5 Ch 1) cAu = c0 = 0.
So, Nul A is closed under scalar multiplication.
Since properties a, b, and c hold, Nul A is a subspace of Rn.
6. a. Complete the statement of the Unique Representation Theorem.
Let β = {v1, v2, … , vn} be a basis for a vector space V. Then for each x in V, there
exists a unique set of scalars c1 , c2 ,, cn such that x  c1v1  c2 v 2    c p v p .
b. Explain why a linearly independent set of four vectors in P3 is a basis for P3. Mention
the appropriate fact(s) from Chapter 4.

2
3

The standard basis for P3 is 1, t , t , t , so dim P3 = 4. By Th. 12 Ch. 4, any set of 4 linearly
independent vectors in P3 is a basis for P3.