MIDTERM 1 SOLUTIONS
1.(a) The columnspace of A is
2
4
8
C(A) = span
,
,
3
6
12
Note that
so
4
2
8
2
=2
and
=4
6
3
12
3
2
C(A) = span
3
and {( 23 )} is a basis for the columnspace.
(b) Perform Gaussian elimination:
8
12
1 2 4
3 6 12
1 2 4
0 0 0
So solutions to A x = 0 are solutions to
2
3
4
6
x1 + 2x2 + 4x3 = 0
0=0
The solutions to this system are
 

 
  
−2x2 − 4x3
−2
−4
x1
 = x2  1  + x3  0 
 x2  = 
x2
1
x3
x3
0
So a basis for the nullspace is
   
−4 
 −2
N (A) =  1  ,  0 


0
1
(c) ( 20 ) does not belong to the columnspace, so there’s no solution to A x = ( 20 ).
( 23 ) is the first column of A, so
 
1
2
A 0 =
3
0
1
2
MIDTERM 1 SOLUTIONS
so the space of solutions to A x = ( 23 ) is the translate of N (A) by
1
0
0
, or

 
 
 
−2
−4
 1

0 + s  1  + t  0  | s, t in R


0
0
1
2. Let V be the span of {u + v, u − v, u + v + w}. V is a subspace of Rn of
dimension at most 3. Note that
1
1
(u + v) + (u − v)
2
2
1
1
v = (u + v) − (u − v)
2
2
w = (u + v + w) − (u + v)
u=
So u, v, and w are in V . By Proposition 12.1, if the dimension of V were less
than 3, then {u, v, w} would be a linearly dependent set. But we know they’re
linearly independent, so the dimension of V must be 3. So by Proposition 12.3,
{u + v, u − v, u + v + w} is a linearly independent set.
3(a) Solutions to rref(A) x = 0 are solutions to
x1 + 3x2 + 2x4 = 0
x3 − 8x4 = 0
x5 = 0
x6 = 0
0=0
Rearranging, we have

 

 
 
x1
−3x2 − 2x4
−3
−2
 x2  

1
0
x
2
  

 
 
 x3  

 
 
8x4
 =
 = x2  0  + x4  8 
 x4  



1
x4
  

0
 
 x5  



0
0
0
x6
0
0
0
A basis for N (A) is
   
−3
−2 





 1   0 








   

0
8
 , 




0   1 






 0   0 






0
0
MIDTERM 1 SOLUTIONS
3
(b) There are pivots in columns 1, 3, 5, and 6 of rref(A), so the corresponding
columns of A give a basis for C(A):
       
4 
3
1
3






 −1 3 1
−1

       
 2  ,  1  , 4 , 1
       


 0  −1 3 1






1
0
1
1
4(a) a = 2 will necessitate

1
2
0
a row interchange. It will go like this:



1 3 2
3 2
R2 R2 −2R1
0 0 −2
6 2
−→
9 5
0 9 5
At this point, the 9 in the second column is going to become a pivot, so it needs to
be moved to the second row.
(b) Row interchange is not always necessary, but it’s optional for any a. If we
row-reduce A, we get






1 3 2
1
3
2
1
3
2
a 6 2 −→ 0 6 − 3a 2 − 2a −→ 0
9
5 
0 9 5
0
9
5
0 6 − 3a 2 − 2a




1
3
2
1 0
1/3

1
5/9  −→ 0 1
5/9
−→ 0
0 6 − 3a 2 − 2a
0 0 −4/3 − a/3
The entry in the lower-right corner is zero when a = −4. When a = −4, the
nullspace is nontrivial. Otherwise, the entry is a pivot and the nullspace is trivial.
(c) As we know from (b), when a = −4, the nullity of A is 1, and so the rank is 2.
Otherwise the nullity is 0, and so the rank is 3.
5(a) Sometimes FALSE.
(b) Always TRUE.
(c) Always TRUE.
(d) Sometimes FALSE.
(e) Sometimes FALSE.
(f ) Sometimes FALSE.
(g) Always TRUE.
(h) Sometimes FALSE.
(i) Always TRUE.
4
MIDTERM 1 SOLUTIONS
(j) Always TRUE.
6(a) If v ∈ V , then v ·0 = 0. So 0 ∈ W .
If w, w0 ∈ W , that means that for any v ∈ V , we have v · w = v · w0 = 0. Then
v ·(w + w0 ) = v · w + v · w0 = 0 + 0 = 0. So w + w0 ∈ W .
If w ∈ W and v ∈ V and c ∈ R, then v ·(c w) = c(v · w) = c0 = 0. So c w ∈ W .
Since W contains 0 and is closed under addition and scalar multiplication, it is
a subspace.
(b) Let


w1
w2 

w=
w3 
w4
w is in W if and only if it’s orthogonal to the basis vectors of V :
 
 
0
1
−3
3

 
w·
−1 , w ·  4 
3
4
This condition is the same as
w1 + 3w2 − w3 + 4w4 = 0
−3w2 + 4w3 + 3w4 = 0
Add the second equation to the first to eliminate w2 :
w1 + 3w3 + 7w4 = 0
−3w2 + 4w3 + 3w4 = 0
Divide the second equation by −3:
w1 + 3w3 + 7w4 = 0
4
w2 − w3 − w4 = 0
3
Solve for w1 and w2 :
w1 = −3w3 − 7w4
4
w2 = w3 + w4
3
So the possible values of w are
 
 
 
w1
−3
−7
w2 
4/3
1
  = w3   + w4  
w3 
 1 
0
w4
0
1
MIDTERM 1 SOLUTIONS
So a basis for W is
5
    
−7 


 −3

4/3  1 
 , 
 1   0 





1
0
7. The system is represented by

1
1

1
0
the augmented matrix

1 1 2 −1
1 1 0 1

0 1 1 1
1 2 4 −3
Row-reduce:

1
0

0
0


1 0
−1

2
0
R2 R2 −R1
R
R
+R
4
3 0
 4 −→
−→
2  R1 R1 +R3 0 −1
R3 R3 −R1
−3
0 0



1 0 1 1
1 0 1 1
1
 R3 12 R4 0 1 0 1
0
1
0
1
−2
R2 −R3 
 −→ 
−→ 
2  R4 R3 0 0 1 3/2
R3 R2 0 0 0 −2
0 0 2 3 −1
0 0 0 −2



1 0
1 0 0 −1/2 3/2
 R4 − 21 R4 0 1
0
1
0
1
−2
R1 R1 −R3 



−→
0 0 1 3/2 −1/2 −→ 0 0
2
0 0 0 −2
0 0


1 0 0 0 1
R1 R1 + 21 R4 ,R2 R2 −R4 0 1 0 0 −1


−→
0 0 1 0 1 
R3 R3 − 23 R4
0 0 0 1 −1
1
0
−1
1
1 2
0 −2
0 −1
2 4

1 1
1
0 −2 2 

0 −1 2 
2 3 −1

1
−2 

−1/2
2
0
0
1
0
−1/2
1
3/2
1

3/2
−2 

−1/2
−1
This represents the solution
x1 = 1
x2 = −1
x3 = 1
x4 = −1
8. A is an m-by-n matrix, so it has at most m pivots. So its rank is at most
m. So by the rank-nullity theorem, its nullity is at least n − m, which is at least
1. So the nullspace of A has dimension at least 1. So the nullspace of A contains a
nonzero vector x, and this vector satisfies A x = 0.
9. The plane is defined by the equation x1 + 2x2 − x3 = 0. Solving for x1 , we
get x1 = −2x2 + x3 . So the vectors in P are the vectors
 
 
 
x1
−2
1
x2  = x2  1  + x3 0
x3
0
1
6
MIDTERM 1 SOLUTIONS
So a basis for P is
    
1 
 −2
 1  , 0


0
1
4
When you throw in the vector 1 , you get a basis for R3 . You can verify this by
1
row-reducing the matrix


−2 1 4
 1 0 1
0 1 1
and seeing that it has rank 3.
Let us give names to these basis vectors:
 
 
 
−2
1
4
x =  1  , y = 0 , z = 1
0
1
1
So every vector w ∈ R3 can be written as a sum a x +b y +c z, with a x +b y ∈ P
and c z ∈ L.
Now suppose it’s also true that w = v1 + v2 for some v1 ∈ P and v2 ∈ L. We
will show that v1 = a x +b y and v2 = c z. Indeed, since v1 is in P , it can be
expressed in terms of our basis for P : That is, v1 = d x +e y for some real numbers
d and e. And similarly, v2 = f z for some real number f . So w = d x +e y +f z.
Now every vector in R3 can be expressed in terms of the basis {x, y, z} in a unique
way; and we have expressed it in two ways,
w = a x +b y +c z
w = d x +e y +f z
So we must have a = d, b = e, c = f . Therefore v1 = d x +e y = a x +b y, and
v2 = f z = c z.
We have shown that any vector x ∈ R3 can be written in one and only one way
as a sum w = v1 + v2 where v1 ∈ P and v2 ∈ L.