Solutions to Math 51 First Exam — April 26, 2012
1. (12 points) Complete the following sentences.
(a) Vectors v and w in R7 are defined to be orthogonal if
(4 points)
v · w = 0.
(b) A basis for a subspace V of Rn is defined to be
(4 points) . . . a linearly independent set of vectors {v1 , . . . , vk } such that
span(v1 , . . . , vk ) = V.
(c) A function T : Rn → Rm is called a linear transformation if
(4 points)
T(x + y) = T(x) + T(y)
T(cx) = cT(x)
for all x, y ∈ Rn , and
for all c ∈ Rn and x ∈ Rn .
Math 51, Spring 2012
Solutions to First Exam — April 26, 2012
Page 2 of 9
2. (10 points) Be careful to answer both parts of the following:
(a) Compute, showing all steps, the reduced row echelon form of the matrix


1 2 0 −1 −2
3 0 6 2
5

A=
1 2 0 −1 −2
2 3 1 1
0
(7 points) First, R2 - 3R1, R3 - R1, R4 - 2R1 gives


1 2 0 −1 −2
 0 −6 6 5 11 


 0 0 0 0
0 
0 −1 1 3
4
Multiplying R4 by -1, exchanging R3 and R4, and dividing R2/6, gives,


1 2 0 −1 −2
 0 −1 1 5/6 11/6 


 0 −1 1 3
4 
0 0 0 0
0
R3 - R2 gives:


1 2 0 −1
−2
 0 −1 1 5/6 11/6 


 0 0 0 13/6 13/6 
0 0 0
0
0
R2 + R3 and 6*R3 gives:


1 2 0 −1
−2
 0 −1 1
3
4 


 0 0 0 13/6 13/6 
0 0 0
0
0
R1 + 2R2, multiplying R2 by -1 and dividing R3 by 13/6 gives:


1 0 2
5
6
 0 1 −1 −3 −4 


 0 0 0
1
1 
0 0 0
0
0
R1 - 5R3, R2 +3R3 gives the RREF:

1
 0

 0
0
(b) Fill in the blanks (no reasoning needed):

0 2 0 1
1 −1 0 −1 

0 0 1 1 
0 0 0 0
Rank of A:
3
Nullity of A:
One point for one answer correct; three points for both answers correct.
2
Math 51, Spring 2012
Solutions to First Exam — April 26, 2012
Page 3 of 9
3. (12 points) Consider the following three points A, B, C in R3 :
A = (1, −1, 3),
B = (4, 1, −2),
C = (−1, −1, 1)
(a) In the triangle ∆ABC, determine the cosine of the angle at vertex B.
−−→
−−→
(4 points) By BA = A − B = (−3, −2, 5) and BC = C − B = (−5, −2, 3), we have
cos(∠B) =
=
=
=
=
−−→ −−→
BA · BC
−−→ −−→
kBAk kBCk
15 + 4 + 15
√
√
2
2
3 + 2 + 52 52 + 22 + 32
34
√ √
38 38
34
38
17
.
19
(b) Let P be the plane in R3 that passes through the points A, B, C. Find a parametric representation
for P .
 

 
 
−3
−5  4

−−→
−−→
(4 points) P = {B + sBA + tBC|, s, t ∈ R} =  1  + s −2 + t −2 s, t ∈ R . Notice


−2
5
3
−−→
−−→
that B can be replaced by any point on the plane ABC, such as A or C. Also, BA and BC can
−−→ −−→
be replaced by any linearly independent vectors v and w in span(BA, BC).
(c) Find an equation for the plane P of part (b), in the form ax + by + cz = d. (Here a,b,c,d are
scalars, and x,y,z are the usual variables for coordinates of points in R3 .)
−−→ −−→
(4 points) The normal vector n = BA × BC = (4, −16, −4). Therefore, the equation is
n · (x, y, z) = n · A,
which becomes x − 4y − z = 2.
Math 51, Spring 2012
Solutions to First Exam — April 26, 2012
Page 4 of 9
4. (16 points) Let


1 0 4 0 −3
A = 0 1 1 0 2  .
0 0 0 1 0
As usual, we’ll write N (A) and C(A), respectively, for the null space and column space of A.
(a) Find, with reasoning, a basis for N (A).
(4 points) Note that A is already in row reduced echelon form, i.e. rref(A) = A. Thus, the free
variables are x3 and x5 . Note that x ∈ N (A) if and only if:
x1 = −4x3 + 3x5
x2 = −x3 − 2x5
x4 = 0.
Thus,
 




 

N (A) = x3 







−4
−1
1
0
0






 + x5 




3
−2
0
0
1








 : x3 , x5 ∈ R.








And a basis for N (A) is given by:














−4
−1
1
0
0
 
 
 
,
 
 
3
−2
0
0
1






 .







3
(b) Find all solutions to the equation Ax =  5 .
−7


3
(4 points) Ax =  5  if and only if:
−7
x1 = −4x3 + 3x5 + 3
x2 = −x3 − 2x5 + 5
x4 = −7.
Hence, the set of solutions is:







N (A) = 







3
5
0
−7
0






 + x3 




−4
−1
1
0
0






 + x5 




3
−2
0
0
1








 : x3 , x5 ∈ R.








Math 51, Spring 2012
Solutions to First Exam — April 26, 2012
Page 5 of 9


1 0 4 0 −3
For quick reference, here again is the matrix:
A = 0 1 1 0 2 
0 0 0 1 0
 
11
0
 

(c) Find a basis for N (A) that contains the vector 
−2, or state why no such basis exists.
0
1


11
 0 



(4 points) Let v = 
 −2 .
 0 
1



 11

1 0 4 0 −3 
 0 




Note that, Av = 0 1 1 0 2
 −2  = 0.
 0 
0 0 0 1 0
1


11
 0 


 ∈ N (A).
−2
Thus, 


 0 
1
Thus, there is a basis for N (A) containing v. Since dim N (A) = 2, v will form a basis for N (A)
along with any other vector in N (A) that is not a scalar multiple of v. Thus, one basis of N (A)
is given by,


 
11 
 −4




  0 

−1


 

 1  ,  −2  .

 




 0   0 






1
0
 
4

(d) Find a basis for C(A) that contains 1, or state why no such basis exists.
0


4
(4 points) Note that dim C(A) = 3. Also C(A) ⊂ R3 . Thus C(A) = R3 . Hence the vector  1 
0
is in C(A) and one possible basis is:
     
1
0 
 4
 1 , 0 , 0  .


0
0
1
Note: In part c and d, 2 points for saying that the vector is in N (A)/C(A) respectively. And
2 more points for finding the basis correctly. The most common mistake in this part is that
students show a different basis and prove that the span contains the concerned vectors. The
problem asks to find a basis containing the vector.
Math 51, Spring 2012
Solutions to First Exam — April 26, 2012
Page 6 of 9
5. (8 points) Let L be the line in R3 that is the intersection of the planes whose equations are
x+y+z =1
x−y+z =1
and
(a) Find L in parametric form.
(4 points) Regard L as the set of solutions to a 2 × 3 linear system as given. The RREF of the
corresponding augmented matrix is
1 0 1
1
0 1 0
0
Therefore, the original two equations are equivalent to the following:
x + z = 1,
y=0
Therefore, the parametric form is given by
{(x, y, z) = (1, 0, 0) + z(−1, 0, 1) : z ∈ R}
 
2
3

(b) Find, with reasoning, a matrix A such that L is the set of solutions to the system Ax = 
4, or
5
state why no such A exists.
(4 points) For this part, we have to reverse the row operations. More precisely, we would like to
create an augmented matrix whose last column is [2 3 4 5]T , with RREF equal to


1 0 1
1
0 1 0
0


0 0 0
0
0 0 0
0
One such augmented matrix is

2
3

4
5
0
1
0
0
2
3
4
5

2
3

4
5
The 4 × 3 submatrix on the left (i.e., the portion consisting of the coefficients) is the A we want.
Math 51, Spring 2012
Solutions to First Exam — April 26, 2012
Page 7 of 9
6. (8 points) Suppose T : R2 → R4 is a linear transformation, and that
 
 
4
8
−1
4
1
0


T
=
=
 1  and T 2
0
1
−2
0
1
0
(a) Find T
and T
.
0
1
(4 points) We have the following two equalities:
1 0
1
1
=
−
and
0
1
2 2
1 0
0
=
.
1
2 2
Since T is a linear transformation, this gives

   
0
4
8





1
1
1
0
−1 1  4  −3

−
=T
T
−
=
=
T

1
0
2
1  2 0   1 
2
1
0
−2
   
4
8
 2
1
1
0
0
4
= 
= T
T
= 
1
2
2
2 0   0 
−1
−2

(b) Let b ∈ R4 . Find one or more conditions on b that determine precisely whether b is equal to
T(x) for some x ∈ R2 ; that is, whether b belongs to im(T). (Your answer should be given in the
form of one or more equations involving the components b1 , b2 , b3 , b4 of b.)


0
4
−3 2 
. The system
(4 points) By part (a), we know that the matrix corresponding to T is 
1
0
1 −1
of equations T(x) = b gives rise to the following augmented matrix, and we find its reduced row
echelon form:




0
4 b1
1 0
b3
 −3 2 b2 
 0 1

−b4 + b3




 1


0 b3
0 0 b1 − 4b3 + 4b4 
1 −1 b4
0 0 b2 + b3 + 2b4
(
b1 − 4b3 + 4b4 = 0,
x
Thus if x =
, then T(x) = b if and only if x = b3 and y = −b4 +b3 and
y
b2 + b3 + 2b4 = 0.
Therefore b is in the image of T if and only if the last two equations above are true.
Note: it actually works to solve this part by using the augmented matrix


4
8 b1
 −1 4 b2 
1
0

 , but the reason this works is that the vectors
and
form a basis
 1
0 b3 
1
2
0 −2 b4
for R2 . A complete solution should state this fact as a justification for the method used.
Math 51, Spring 2012
Solutions to First Exam — April 26, 2012
Page 8 of 9
7. (10 points) Each of the statements below is either always true (“T”), or always false (“F”), or sometimes true and sometimes false, depending on the situation (“MAYBE”). For each part, decide which
and circle the appropriate choice; you do not need to justify your answers.
1
0
2
In all these statements, the vector e1 =
(in R ), and similarly e2 =
.
0
1
(a) Given a 2 × 5 matrix A, the equation Ay = e1 has no solutions y in R5 .
T
F
MAYBE
(Equivalently, the issue is whether or not e1 lies in C(A).) It might be that A has rank 2, i.e.
that C(A) = R2 , in which case the equation has a solution; however, if A is the zero matrix,
then C(A) contains only the zero vector, and the above equation does not have a solution.
(b) Given a 5 × 2 matrix B, the equation Bz = Be1 has infinitely many T
F
MAYBE
solutions z in R2 .
The equation has at least one solution, z = e1 , but whether it has infinitely many is equivalent
to whether N (B) is non-trivial, i.e., whether B has positive nullity. This depends, since B could
have rank 2 (and nullity 0); but if B is the zero matrix, then B has rank 0 (and nullity 2).
(c) Given vectors v1 , v2 , v3 in R2 with the property that each of the sets
{v1 , v2 },
{v2 , v3 },
T
F
MAYBE
and {v1 , v3 }
is linearly independent, the set {v1 , v2 , v3 } is also linearly independent.
No set of three vectors in R2 is ever linearly independent, since dim(R2 ) = 2 (see Prop. 12.1).
(d) Given vectors w1 , w2 , w3 in R5 with the property that each of the sets
{w1 , w2 },
{w2 , w3 },
T
F
MAYBE
and {w1 , w3 }
is linearly independent, the set {w1 , w2 , w3 } is also linearly independent.
For example, let w1 = (1, 0, 0, 0, 0) and w2 = (0, 1, 0, 0, 0). On the one hand, consider w3 =
w1 +w2 , in which case the three two-element sets are each independent but the three-element set
is not; on the other hand, consider w3 = (0, 0, 1, 0, 0), in which case all the sets are independent.
(e) Given nonzero v ∈ R2 , the set
T
F
MAYBE
V = {x ∈ R2 | kx + vk2 = kxk2 + kvk2 } is a subspace of R2 .
The key point is that V is exactly the set of those vectors orthogonal to v: this is because
kx + vk2 = (x + v) · (x + v) = x · x + x · v + v · x + v · v = kxk2 + kvk2 + 2(x · v),
which equals kxk2 + kvk2 if and only if x · v = 0. (Aside: compare with Prop. 4.7.) Now,
(i) 0 ∈ V because 0 · v = 0; and
(ii) if x1 , x2 ∈ V , then (x1 + x2 ) · v = x1 · v + x2 · v = 0 + 0 = 0, so x1 + x2 ∈ V ; and
(iii) if x ∈ V and c is a scalar, then (cx) · v = c(x · v) = c(0) = 0, so cx ∈ V .
Thus, V is a subspace of R2 .
Math 51, Spring 2012
Solutions to First Exam — April 26, 2012
(f) Given nonzero w ∈ R2 , the set
Page 9 of 9
T
F
MAYBE
W = {x ∈ R2 | kx + wk = kxk + kwk} is a subspace of R2 .
First note that w lies in W , because kw + wk = k2wk = 2kwk = kwk + kwk. But also −w does
not lie in W , since k − w + wk = 0 6= k − wk + kwk = 2kwk since w is nonzero; thus, W is not
closed under scalar multiplication and so cannot be a subspace. (For a more geometric reason,
consider the fact that by the Triangle Inequality, also known as Prop. 4.4, kx + wk = kxk + kwk
holds if and only if x = cw for some positive scalar c. This means that W is a ray, or half-line,
which is not a subspace for the same reason as before.)
(g) Given a counterclockwise rotation Rotθ : R2 → R2 through an angle θ,
the set {Rotθ (e1 ), Rotθ (e2 )} is a basis for R2 .
T
F
MAYBE
The key to this part, and to parts (h)-(j) below, is determining whether the given set is linearly
independent: we know that since dim(R2 ) = 2, any set of two vectors in R2 will be a basis for
R2 if and only if the set is linearly independent (see Prop. 12.3). (Furthermore, we know that
any set of two vectors is independent if and only if the two vectors are not collinear.)
In the case of {Rotθ (e1 ), Rotθ (e2 )}, note that since the orthogonal vectors e1 and e2 are being
rotated by the same angle θ, the images will still be orthogonal, and therefore not possibly
collinear. Thus, the set must be independent and thus a basis for R2 .
(h) Given a counterclockwise rotation Rotθ : R2 → R2 through an angle θ,
the set {e1 , Rotθ (e1 )} is a basis for R2 .
T
F
MAYBE
The vectors e1 and Rotθ (e1 ) might be collinear, if θ is a multiple of π, or they might not be
collinear (for any other value of θ, such as θ = π/2). So the set {e1 , Rotθ (e1 )} might be linearly
dependent (and therefore not a basis), or it might be independent (and therefore a basis).
(i) Given a projection ProjL : R2 → R2 onto a line L containing the origin,
the set {ProjL (e1 ), ProjL (e2 )} is a basis for R2 .
T
F
MAYBE
The vectors ProjL (e1 ) and ProjL (e2 ) both lie in L, so they must be collinear. Thus, the set
{ProjL (e1 ), ProjL (e2 )} is always linearly dependent, and therefore cannot be a basis.
(j) Given a reflection ReflL : R2 → R2 through a line L containing the origin,
the set {ReflL (e1 ), ReflL (e2 )} is a basis for R2 .
T
F
MAYBE
The set is linearly independent, and therefore forms a basis, because the vectors ReflL (e1 ) and
ReflL (e2 ) are not collinear. (There are many ways to see this: for example, if the reflected vectors
were collinear, then reflecting each of them a second time would still leave them collinear. But
reflecting twice has the effect of doing nothing at all, which means that we’d be saying that e1
and e2 are collinear – and this is false. Alternatively, you can use the formula for ReflL to show
that the reflected vectors ReflL (e1 ) and ReflL (e2 ) are in fact still orthogonal [and nonzero],
which means they cannot be collinear.)