Math 314 - Fall 2011 - Instructor: Alexandra Seceleanu
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Practice Midterm 1 - Solutions
1.
(a) Find the equation of the plane in R3 that passes through the origin
and has normal vector
 
1
~n = 2 .
3
 
0
Solution: The normal equation ~n · ~x = ~n · ~p, where ~p = 0 is the
0
position vector of the origin yields
x + 2y + 3z = 0
(b) Find a and b such that the plane in R3 of equation ax + by + z = 2 is
parallel to the plane in part a).

 
 
 
 
a
1
a
c
a=c
0








Solution: ~n = b = c 2 =⇒ b = 2c =⇒ b = 2c =⇒

1
3
1
3c
1 = 3c

 c = 1/3
a = 1/3

b = 2/3
(c) Find the vector equation of the line in R3 which passes through the
point P = (1, 1, 1) and is perpendicular to the plane in a).
 
1
~

Solution: The vector equation ~x = td + ~p, where ~p = 1 is the posi1
 
1
tion vector of the point P and ~d = 2 is the direction perpendicular
3
to the plane in part a) yields
    

1
1
t+1
~x = t 2 + 1 = 2t + 1 , t ∈ R.
3
1
3t + 1
2. Find all the values of k such that the linear system

= 0
 y + 2kz
x + 2y + 6z = 2

kx + 2z
= 1
has
(a) no solutions.
(b) a unique solution.
(c) infinitely many solutions.
Solution: The augmented matrix of the system row reduces as follows:


0 1 2k | 0
1 2 6 | 2
k 0 2 | 1

R1 ↔R2
−→
R30 =R3 −kR1
−→
R30 =R3 +2kR2
−→
1 2 6 |
0 1 2k |
k 0 2 |

1 2
6
|
0 1
2k
|
 0 −2k 2 − 6k |
1 2
6
0 1
2k
0 0 2 − 6k + 4k 2

2
0
1

2
0 
1 − 2k 
|
2
|
0 
| 1 − 2k
Note that in order to row-reduce further we need to know whether the
entry 2 − 6k + 4k 2 is zero. Also note 2 − 6k + 4k 2 = 2(2k − 1)(k − 1).
Case 1: if k 6= 1/2, k 6= 1 then the coefficient matrix is invertible because
it has rank 3. It follows by the theorem of invertible matrices that the system
has unique solution.
Case 2: if k = 1/2 then the last variable is free and therefore there are
infinitely many solutions.
Case 3: if k = 1 then the last row gives the equation 0 = −1 which says
the system is inconsistent (no solutions).
2

3.

1
5 −3
1 −2 .
(a) Find the inverse of the matrix A =  0
−2 −7 0
Solution: We row reduce the multi-augmented matrix [A|I3 ].


1
5 −3 | 1 0 0
0
1 −2 | 0 1 0
−2 −7 0 | 0 0 1
R30 =R3 +2R1
−→
R30 =R3 −3R2
−→
R10 =R1 −5R2
−→

1
0
0
1
0
0
1
0
0
5
1
3
5
1
0
0
1
0
−3
−2
−6
−3
−2
0
7
−2
0

| 1 0 0
| 0 1 0
| 2 0 1
| 1 0 0
| 0 1 0
| 2 −3 1
| 1 −5 0
| 0 1 0
| 2 −3 1
Since A does not row-reduce to I3 , A does not have an inverse.
(b) Check that the matrix you found is really A−1 .
Solution: Nothing to do. Trick question.
 
−4
(c) Use part a) to write ~b = −1 as a linear combination of the vectors
5  
 
 
1
5
−3
~v1 =  0  , ~v2 =  1  , ~v3 = −2.
−2
−7
0
Solution: By part a) we know that a basis for Span{~v1 , ~v2 , ~v3 } is given
by the first two vectors, so it’s enough to solve the system c1~v1 +c2~2 = ~b
whose augmented matrix is






1
5 | −4 0
1 5 | −4
1 5 | −4
R =R3 +2R1
3 −3R2
1 −5R2
0
0 1 | −1 R−→
0 1 | −1 R−→
1 | −1 3 −→
−2 −7 | 5
0 3 | −3
0 0 | 0
Hence c1 = 1, c2 = −1 and the required linear combination is
~b = 1~v1 + (−1)~v2 + 0~v3 .
3


1 0 | 1
0 1 | −1
0 0 | 0

1
 1
4. Let A = 
 1
1
2 1 0
1 0 1
0 −1 1
1 0 1


1

1 
 . You may assume that RREF (A) = 

0 
1
1
0
0
0

0 −1 0 −1
1 1 0 1 
.
0 0 1 1 
0 0 0 0
(a) Find a basis for the row space of A.
Solution:
A basis for the row space of A is given by the rows in R containing
leading 1’s:
{ 1 0 −1 0 −1 , 0 1 1 0 1 , 0 0 0 1 1 }.
(b) Find a basis for the null space of A.
Solution:
We compute the nullspace of A by solving the system A~x = ~0:



1 2 1 0 1 | 0
1 0 −1 0 −1 | 0
 1 1 0 1 1 | 0  RREF  0 1 1 0 1 | 0


A=
 1 0 −1 1 0 | 0  −→  0 0 0 1 1 | 0
1 1 0 1 1 | 0
0 0 0 0 0 | 0
x3 = s and x5 = t are free variables. The nullspace is



s+t









−s
−
t






s  , s, t ∈ R
null(A) =
=




 −t 






t
   
1 

 1








−1 −1

  
= Span 
 1  ,  0  & a


 0  −1






0
1




  

 
1
1








−1

−1
 

 
 




s  1  + t  0  , s, t ∈ R




0
−1






0
1   
1 

 1








−1 −1

  
basis is given by 
 1  ,  0 


 0  −1






0
1
(c) Check that the rank theorem holds for A.
Solution: From a) rank(A) = 3, from b) nullity(A) = 2 and 2 +
3 = 5 = # of columns (this verifies the rank theorem: rank(A) +
nullity(A) = # of columns of A).
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5.
(a) Define what a subspace is.
Solution:
A set S is a subspace of Rn if it satisfies:
(a) ~0 ∈ S
(b) for any two vectors ~u, ~v ∈ S =⇒ ~u + ~v ∈ S
(c) for any vector ~u ∈ S and any scalar c ∈ R =⇒ c~u ∈ S
x
(b) Is S = {
with x ≥ 0, y ≤ 0} a subspace of R2 ? Why or why not?
y
Solution:
S is not a subspace or R2 because
it is not closed under scalar
multipli
−1
1
cation. For example if ~u =
∈ S and c = −1 then c~u =
6∈ S.
−1
1
6. Are the following statements true or false. If you believe a statement is
false give a counterexample. If you believe the statement is true state why
it is true.
(a) Let A be a 3 × 4 matrix. Then the columns of A must be linearly
dependent vectors.
Solution: True. The rows are 4 vectors in R3 and since 4¿3 they will be
linearly dependent by a theorem we learned (it is denoted by Theorem
2.8 in the textbook and class notes). Another way to see this is that
the rank of A is at most 3, hence the rank of A is less than the number
of vectors (4).
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(b) Let A be a 3 × 4 matrix. Then the nullity of A must be 1.


1 0 0 0
0 1 0 0

Solution: False. For A = 
0 0 0 0, rank(A) = 2 and since
0 0 0 0
rank(A) + nullity(A) = 4 we have nullity(A) = 2.
(c) If A, B, C are all n × n matrices and AB = AC then B = C
Solution: False. Let A be the zero matrix and B 6= C any two distinct
matrices. Then AB = AC = On×n but B 6= C.
(d) If A, B are n×n matrix and AB = On×n , then A = On×n or B = On×n .
(Here On×n is the n × n zero matrix.)
0 1
Solution: False. Consider A = B =
. Then AB = On×n but
0 0
neither A nor B are the zero matrix.
7. Prove that if A is an invertible n × n matrix then
(AT )−1 = (A−1 )T .
(Recall AT means the matrix transpose and A−1 means the matrix inverse.)
Solution: All we need to show is that
(A−1 )T · AT = In and AT · (A−1 )T = In
(A−1 )T · AT = (A · A−1 )T = (In )T = In
AT · (A−1 )T = (A−1 A)T = (In )T = In
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