Linear Algebra MA237 Section 1
Fall 2009 Dr. Byrne
Homework for Section 2.3
Assigned 09/30
Due 10/5
1. We’ve learned two different ways of finding the basis for the span of a set of vectors.
a. Write down 4 vectors that belong to 3 (your choice) and label them v1, v2, v3
and v4. (They should be 3x1 vectors).
4
v1 = 7  , v2 =
9 
14
17 , v =
  3
19
0 
0  , v =
  4
0 
4
4
 
 4 
b. Let C be the 3x4 matrix in which the columns are equal to your four vectors.
Row reduce this matrix to determine which of your original 4 vectors form a
basis for the span of v1, v2, v3 and v4.
4 14 0 4
1 0 0  0.4


C  7 17 0 4 , rref (C )  0 1 0 0.4 
9 19 0 4
0 0 0
0 
The vectors that form a basis for the span of {v1, v2, v3, v4} are {v1, v2}.
c. Let R be the 4x3 matrix in which the rows are equal to your four vectors. Row
reduce this matrix to determine which of your original 4 vectors form a basis
for the span of v1, v2, v3 and v4.
1 0  2 
4 7 9
3

14 17 19
0
1
5
/
3


 , rref (R ) 
R
0 0
0 0 0
0 




0 
4 4 4
0 0
The vectors that form a basis for the span of {v1, v2, v3, v4} are {v1, v2}.
d. Compare the basis obtained in part (b) with the basis obtained in part (c).
Even if the bases are not identical they are equivalent since they span the same
subspace of 3. Comparing the methods in part (b) and part (c), which did
you prefer for obtaining the basis?
I prefer the column-method because you can easily see how the 3rd and 4th
vectors are linear combinations of the 1st and 2nd from rref(C), whereas seeing
how the third column of R is related to the first 2 is not very useful.
2. Let A be an m x n matrix. If n>m, then there are more variables than equations and
either there are infinitely many solutions or the system is inconsistent.
a. Write down a linear system Ax=B of your choice in which n>m. (Note that
you need to make up numbers for the entries in A and make up numbers for
the entries in B, and B will be an m x 1 column vector.) Find the set of
solutions to your linear system.
 x1 
x 
2
1 3 2 6 15    1

Ax  B : 
x
 3   
2 4 4 8 20   1
 x4 
 x5 
 1 3 2 6 15 1  1 0 2 0 0  1 2 


Solutions: rref  
 2 4 4 8 20 1  0 1 0 2 5 1 
 

2 
 x1    1 2  2u   1 2 
  2  0   0 
 

x   1
 0    2    5
1
 2   2  2t  5s   2 
     



 x3   
 1   t  0   s 0 


u
u
  0 
  
     
  0 
t
 x4  
0  1  0
 

 x5  
 0   0   1 
s

  0 
b. The set of solutions to your linear system was either empty or infinite. Can
you figure out how to modify your system by changing some elements in A or
changing B to switch to the other case? (Hint: remember that a system is
consistent IFF B belongs to the column space of A.)
The solution set was infinite, with 3 free variables. How can I change A or B
so that there is no solution (empty solution set)?
… I cannot just pick a different B, because the column space of A spans 2
and thus I will have solutions for any B I choose. I need to change A so that it
doesn’t span 2, and, in particular, doesn’t have B in its span. So I would
modify A so all the columns are linearly dependent:
1 1 1 1 1 0
1 1 1 1 1 
A : 
. Now, rref(A|B)= 


 2 2 2 2 2
0 0 0 0 0 1
which yields the inconsistent equation 0=1 in the second row.
3. Let A be an m x n matrix. If m>n, then there are more equations than variables and
there are infinitely many solutions, exactly one solution or the system is inconsistent.
a. Write down a linear system Ax=B of your choice in which m>n. Find the set
of solutions to your linear system.
1 11
1
x


Ax  B : 2 5   1   1
x
3 4   2  1
 1 11 1   1 0 1 

  

Solutions: rref  2 5 1    0 1 0 
 
   

 3 4 1   0 0 1 
This linear system is inconsistent (and the solution set is empty) because we have
0=1 in the last row.
b. The set of solutions to your linear system was either empty, had exactly one
solution, or was infinite. Can you figure out how to modify your system by
changing some elements in A or changing B to switch to the other cases?
Explain your reasoning for up to 2 bonus points.
Oops… I forgot I was offering bonus points for these. I will need to recollect
the homework!