EXAM
Exam #2
Math 2360
Summer II, 2000
Morning Class
Nov. 15, 2000
ANSWERS
i
60 pts.
Problem 1. Consider the matrix

1 6
4

 0 −1 −1


1
A=
 0 1

 2 1 −3

1 0 −2
The RREF of A is the matrix

1 0

 0 1


R=
 0 0

 0 0

0 0
−1
19
0
−3
−2
7
3
−5
2
−5
−2
0
−1
1
0
3
0
1
−2
0
0
0
0
0
0

5

0 


6 
.

−5 

−4
2


0 


−3 
.

0 

0
A. Find a basis for the nullspace of A.
Answer :
Solve the homogeneous system Ax = 0. The system Rx = 0 has the same
solutions. Call the variables x1 , x2 , . . . , x6 . Looking at R, we see that x1 , x2
and x4 are leading variables and the rest are free variables. Set x3 = α,
x5 = β and x6 = γ. Working up from the bottom of the matrix we get
x4 − 2x5 − 3x6 = 0 =⇒ x4 = 2x5 + 3x6 = 2β + 3γ
x2 + x3 + 3x5 = 0 =⇒ x2 = −x3 − 3x5 = −α − 3β
x1 − 2x3 − x5 + 2x6 = 0 =⇒ x1 = 2x3 + x5 + x6 = 2α + β + γ
This gives the following parameterization of the nullspace
  

 
 
 
x1
2α + β + γ
2
1
1
x2   −α − 3β 
−1
−3
0
  

 
 
 
x3  

 
 
 
α
 =
 = α  1  + β  0  + γ 0
x4   2β + 3γ 
0
2
3
  

 
 
 
x5  

0
1
0
β
x6
γ
0
0
1
Thus, a basis for the nullspace of A is
 
 
1
1
−3
−3
 
 
0
0
 ,
 ,
2
2
 
 
1
1
0
0
1
 
1
0
 
0
 .
3
 
0
1
B. Find a basis for the rowspace of A.
Answer :
A basis for the rowspace of A is given by the nonzero rows of the RREF.
Hence a basis for the rowspace of A is
1 0 −2 0 −1 2 ,
0 1 1 0 3 0 ,
0 0 0 1 −2 −3 .
C. Find a basis for the columnspace of A
Answer :
The leading entries in R are in columns 1, 2 and 4. A basis of the columnspace
of A is formed by the corresponding columns of A. Thus, a basis for the
columnspace of A is
 
 
 
6
−1
1
0
−1
0
 
 
 
−2 .
 1 ,
0 ,
 
 
 
1
3
2
2
1
0
50 pts.
Problem 2. Let A be a 5 × 6 matrix and let B be a 4 × 4 matrix.
A. What is the largest possible value of the rank of A?
Answer : The rank of A is defined as the value of two equal quantities:
the dimension of the rowspace of A and the dimension of the
columnspace of A. Thus, the rank must be less than or equal
to the number of rows and less than or equal to the number of
columns. In the present case, the largest possible value for the
rank is 5.
B. If the nullspace of A has dimension 3, what is the rank of A?
Answer : The rank of A plus the nullity of A is equal to the number of
columns in A. The nullity is the dimension of the nullspace.
Thus, we must have rank + 3 = 6, so the rank of A is 3.
C. If the rowspace of B has dimension 1, what is the dimension of the nullspace
of B?
Answer : The dimension of the rowspace is the same as the rank of B. The
rank plus the nullity is equal to the number of columns. Thus,
we have 1 + nullity = 4, so the nullity (which is the dimension
of the nullspace) is 3.
2
40 pts.
Problem 3. Consider the following three vectors in R4 .
 
 
 
2
3
1
1
3
0



v1 = 
v2 = 
v3 = 
1 ,
1 ,
1
1
1
2
In each part, determine if the given vector is in span(v1 , v2 , v3 ). If so, express
it as a linear combination of v1 , v2 and v3 .
A.


2
−1

w1 = 
2
3
B.
 
7
5

w2 = 
3
6
Answer :
First put the vectors v1 , v2 , v3 , w1 and w2 into a matrix A. Thus,


2 3 1 2 7
1 3 0 −1 5

A=
1 1 1 2 3 .
1 1 2 3 6
The RREF of A is

1
0
0
2

 0

R=
 0

0
1
0
−1
0
1
1
0
0
0
0


0 

.
0 

1
In R we have col4 (R) = 2 col1 (R) − col2 (R) + col3 (R). The columns of A
have the same relation. Expressing this in terms of the original vectors, we
see that
w1 = 2v1 − v2 + v3 ,
3
so w1 ∈ span(v1 , v2 , v3 ).
In R, the last column is not in the span of the first three columns (any linear
combination of the first three columns would have a zero in the bottom
row). Thus, the same holds for the columns of A. We conclude that w2 ∈
/
span(v1 , v2 , v3 ).
Alternatively, one can view the one in the fourth row of the last column of
R as showing that the system Ax = w2 is inconsistent, so w2 is not a linear
combination of the columns of A, and so it’s not a linear combination of
v1 , v2 , v3 .
40 pts.
Problem 4. Determine if the following vectors in R4 are linearly independent.
Justify your answer.
 
 
 
3
4
1
4
5
2





v1 =   ,
v2 =   ,
v3 =  
7
9
4
3
3
1
Answer :
Put these vector in a matrix A, so

3
4
A=
7
3
4
5
9
3

1
2
.
4
1
1
0
0

 0

R=
 0

0
1

0 

.
1 

0
The RREF of A is

0
0

The columns of R are independent, so the same must be true of the columns of
A. Thus, v1 , v2 and v3 are independent.
Another way of looking at this computation is to regard R as saying there
are no free variables in the system Ax = 0, so this system has only the trivial
solution. Thus, the columns of A satisfy only the trivial linear relation, and so
are independent.
4
60 pts.
Problem 5. Pare the following set of vectors down to a basis of R3 . Express
the vectors that are not in the basis as linear combinations of the
basis vectors.
 
 
 
 
 
3
2
−4
1
2
v1 = 2 ,
v2 = 2 ,
v3 = −2 ,
v4 = 1 ,
v5 = 3
2
1
−3
1
1
Answer :
Put the vectors in as the columns of a

3 2

A=
 2 2
2 1
matrix A, so

−4 1 2

−2 1 3 
.
−3 1 1
The RREF of A is

1
0
−2
0

R=
 0
1
1
0
0
0
1
0
−1


2 
.
1
The leading entries of R are in columns 1, 3 and 4. Thus, the corresponding
columns of A form a basis for the column space of A, which is the same as
S = span(v1 , v2 , v3 , v4 , v5 ). Thus, v1 , v2 and v4 form a basis of S. Since S is
a 3-dimensional subspace of R3 , S = R3 . Thus, v1 , v2 and v4 are a basis of R3 .
We need to express v3 and v5 as linear combinations of our basis vectors.
We can read off these linear relationships from the columns of R. Thus, we have
col3 (R) = −2 col1 (R) + col2 (R), so we must have v3 = −2v1 + v2 . Similarly,
col5 (R) = − col1 (R) + 2 col2 (R) + col4 (R), so we have v5 = −v1 + 2v2 + v4 .
40 pts.
Problem 6. The two matrices


2 1 −1
A = 1 1 0  ,
1 2 1

1 0
B = 0 1
0 0

−1
1
0
are row equivalent, so they have the same rowspace. Do they have the same
columnspace? Explain.
Answer :
As many people pointed out, row equivalent matrices need not have the same
column space. But, in a particular example, they may or may not have the
5
same column space—you have to
the matrices
1
0
check your particular matrices. For example,
0
,
1
0
1
1
0
are row equivalent and have the same columnspace (namely R2 ).
A quick reference to a calculator shows that B is the RREF of A. This gives
the vectors
 
 
2
1
a1 = 1 ,
a2 = 1
(1)
1
2
as a basis of the columnspace of A and the vectors
 
 
1
0
e1 = 0 ,
e2 = 1
0
0
(2)
as a basis for the columnspace of B. Certainly these are different lists of vectors,
but what we want to compare are the spans of these sets of vectors. It is certainly
possible to have span(v1 , v2 ) = span(w1 , w3 ) even though v1 , v2 and w1 , w2
are different lists of vectors (think of different bases for R2 ). Thus, merely
noting that we have different lists of vectors in (1) and (2) is not enough to
show that the columnspaces are different; we have to consider the particular
vectors involved.
If the columnspaces of the matrices where equal, we would have
span(a1 , a2 ) = span(e1 , e2 ).
If this were true, we would have (in particular) a1 ∈ span(e1 , e2 ). If this is true,
we would be able to write a1 as a linear combination of e1 and e2 , i.e., we would
have a1 = αe1 + βe2 for some scalars α and β. Then we would have
 
 
   
2
1
0
α
1 = α 0 + β 1 = β  .
1
0
0
0
But, this is impossible, because the right-hand side will always have a zero in
the bottom row.
Thus, a1 is not in the columnspace of B, so the columnspaces of the two
matrices are not equal.
6