Math 3323, Spring 2011 Exam 2 Solutions
1. Let u  (0,2,5) , v  (1,3,1) and w  (3,0,0) . Find
a) (3 pts) u  2v  69
b)
(3 pts) proj u v  (0,2 / 29,5 / 29)
c) (3 pts) A unit vector in the same direction of w  v  (4 / 26 ,3 / 26 ,1 / 26 )
d) (3 pts) u  2v  (34,10,4)
e) (4 pts) The area of the triangle with vertices u, v and w.
uv  (1,5,4) & vw  (4,3,1)
A  1 / 2 uv  vw  867 / 2  17 3 / 2
2. Consider the plane through the point P(1,2,3) with normal vector n  (1,4,3) , find
a) (4 pts) An equation for this plane:  x  4 y  3z  0
b) (4 pts) The distance from the plane to the point Q(7,3,4)
ax0  by 0  cz 0  d
7
D

26
a2  b2  c2
c) (4 pts) Parametric equation for the line passing through Q perpendicular to the
plane
x  x0  at  7  t
y  y 0  bt  3  4t
z  z 0  ct  4  3t
3. (6 pts) Prove that if
orthogonal.
uv
2
are vectors in R n and u  v
2
 u  v
2
2
then
 (u  v)  (u  v)  (u  u )  (u  v)  (v  u )  (v  v)  u  2(u  v)  v
2
 2(u  v)  0  (u  v)  0  u and v are orthogonal
4. (4 pts) If u  v  0 what can you same about
? u & v are parallel
5. (6 pts) Find the equation of the line which is the intersection of the planes
x  4 y  3z  0 and 2x  2z  8
1  4 3 0 
1  4 3 0 
1  4 3 0 
 2 0  2 8   0 8  8 8   0 1  1 1 






zt
y  1 t  y  x  3
x  4t
are
2
6. (6 pts) Let W denote the subset M 2x 2 consisting of all 2x2 matrices A such
that det( A)  0 . Is W a subspace of M 2x 2 ?
a b 
e
det( A)  ad  bc  0 , B  
A

c d 
g
f
det( B)  eh  fg  0
h 
det( A  B)  (a  e)( d  h)  (b  f )(c  g )
det( A  B)  ad  ed  ah  eh  bc  bg  fc  fg
det( A  B)  ed  ah  bg  fc  0
Since the determinant does not equal zero it is not closed under addition.
ka kb
kA  
det( kA)  k 2 ad  k 2 bc  k 2 (ad  bc)  0

 kc kd 
It is closed under scalar multiplication.
Since it is not closed under addition, it is not a subspace.
a  e b  f 
A B  

c  g d  h 
7. (6 pts) Consider the vectors (2,-1,3), (4,1,2) and (5,-1,8). Are these vectors linearly
independent? Do these vectors span R 3 ? Do these vectors form a basis for R 3 ? Explain.
2 4 5
A   1 1  1 det( A)  15  0
 3 2 8 
By equivalence statements Ax  b is consistent for all b, therefore vectors span R 3 . Also
by equivalence statements Ax  0 only has trivial solutions, therefore vectors are linearly
independent. Spans and linearly independent so yes it forms a basis for R 3 .
8. (12 pts) Consider the set of all ordered pairs of real number (i.e. R 2 ) with the following
operations of “addition” and “scalar multiplication”:
(u1 , u2 )  (v1 , v2 )  (2u1  u2 , v1  v2 ) and k (u1 , u2 )  (ku1 ,2ku2 ) .
Determine if the following are true. If so, you need to prove the axiom is true through a
short calculation. Show all work.
a) Is addition commutative (Axiom 2)? No
(u1 , u2 )  (v1 , v2 )  (2u1  u2 , v1  v2 )
(v1 , v2 )  (u1 , u2 )  (2v1  v2 , u1  u2 )
u v  vu
b) Is there a zero vector (Axiom 4)? No
(u1 , u2 )  (a, b)  (2u1  u2 , a  b)  (u1 , u2 )
There is no (a,b) that would satisfy this, so no zero vector.
c) Does k ((u1 , u2 )  (v1 , v2 ))  k (u1 , u2 )  k (v1 , v2 ) (Axiom 7)? No
k ((u1 , u 2 )  (v1 , v 2 ))  (2ku1  ku2 ,2kv1  2kv2 )
k (u1 , u 2 )  k (v1 , v 2 )  (2ku1  2ku2 , kv1  2kv2 )
9. Consider the subset of polynomials, {1  x, x  2 x 2 ,1  x 2 } in the vector space P2 .
a) (5 pts) Do the polynomials form a linearly independent subset of P2 ? Explain.
1 x
W  1
0
x  2x 2 1  x 2
1  4x
2 x  2
4
2
Wronskian is not zero therefore they are linearly independent.
b) (5 pts) Do the polynomials span P2 ? If so, write 2  3x  x 2 as a linear
combination of the given polynomials.
Yes they span because the dimension of P2 is 3 and we have 3 linearly
independent polynomials.
k1 (1  x)  k 2 ( x  2 x 2 )  k 3 (1  x 2 )  2  3x  x 2
k1  k 3  2
k1  7
 k1  k 2  3 Set up as a matrix and solve. k 2  4
2k 2  k 3  1
k3  9
10. Suppose that x1  1, x2  2, x3  4, x4  3 is a solution of a non-homogeneous linear
system Ax  b and that the solution set of the homogeneous
system Ax  0 is x1  3r  4s, x2  r  s, x3  r , x4  s .
a) (3 pts) Find the vector form of the general solution of Ax  0 .
 3  4 
 1   1
 r    s 
1 0
   
0 1
b) (5 pts) Find the vector form of the general solution of Ax  b .
  1  3  4 
 2   1   1
    r    s 
4 1 0
     
 3  0   1 
x y z 0
3 x  2 y  2 z  0

11. Given the following linear system 
4 x  3 y  z  0
6 x  5 y  z  0
a. (5 pts) Find the solution space of the homogeneous linear system and find the
dimension of that space.
1 1 1 0 
1 1 1 0
zt
4
 3 2  2 0



  0 1 5 0 
 t  5
y  5t
 4 3  1 0
0 0 5 0 
 1 
x  4t




6 5 1 0
0 0 5 0 
There is one free variable so the dimension is one.
b. (2 pts) Find a basis for the nullspace of A
4
 5
 
 1 
c. (2 pts) Find a basis for the row space of A
1 1 1, 0 1 5
d. (2 pts) Find a basis for the column space of A
1  1 
3 2
 ,  
4 3
   
6  5 
e. (2 pts) Find rank(A) and nullity(A)
rank(A) = 2 and nullity(A) = 1
Extra Credit: Let W denote the subset M 2x 2 consisting of all symmetric 2x2 matrices.
(Note W is a subspace of M 2x 2 . Find a spanning set for W. What is the dimension of W?
a c 
1 0
0 0  0 1 
V 
 a
 b


  c

 c b
0 0 
0 1 1 0
1 0 0 0 0 1 
Spanning Set = 
, 
, 
  . Note that the linear combination of these three
0 0 0 1 1 0 
linearly independent matrices forms all symmetric matrices. The dimension is three.