Download Solutions to Math 51 First Exam — October 13, 2015

Solutions to Math 51 First Exam — October 13, 2015
1. (8 points) (a) Find an equation for the plane in R3 that contains both the x-axis and the point
(0, 0, 2). The equation should be of the form ax + by + cz = d. Show your work.
Since the plane contains x-axis, it contains origin, so d = 0. And for any point (t, 0, 0) on x-axis,
at = 0, so a = 0.
Since (0, 0, 2) is on the plane, 2c = 0 and hence c = 0.
So the plane is
by = 0, b 6= 0
which is equivalent to
y = 0.
(b) Let P be the plane through the origin in R3 spanned by the vectors
 
 
1
2
1 and 3 .
2
2
Let Q be the plane parallel to P , passing through the point (1, 1, 1). Write down an equation for
the plane Q in the form ax + by + cz = d. Show your work.
Since (1, 1, 1) is on the plane, we can let the plane to have form
a(x − 1) + b(y − 1) + c(z − 1) = 0.
And since the normal vector n = (a, b, c) is orthogonal to (1, 1, 2) and (2, 3, 2), we have
a + b + 2c = 0
and
2a + 3b + 2c = 0.
So we can solve n = (a, b, c) up to a scalar.
a = −4, b = 2, c = 1
So the plane is
−4(x − 1) + 2(y − 1) + (z − 1) = 0.
Rewrite this as
−4x + 2y + z = −1.
Math 51, Autumn 2015
Solutions to First Exam — October 13, 2015
Page 2 of 14
2. (10 points) Consider the 4 × 4 matrix A:

1
1
A=
1
1
2
2
2
2
1
2
4
3

3
4
.
6
5
(a) Find, showing all of your work, a basis for N (A), the null space of A.
Find rref(A):

1
1
A=
1
1
2
2
2
2
1
2
4
3


1
3


0
4
→
0
6
0
5
2
0
0
0
1
1
3
2


1
3


0
1
→
0
3
0
2
2
0
0
0
0
1
0
0

2
1
 = rref(A)
0
0
Since N (rref(A)) = N (A), we obtain the following system of equations:


x1 + 2x2 + 2x4 = 0


x + x
=0
3
4

0
=0



0
=0
Writing the pivot variables x1 , x3 in terms of the free variables x2 , x4 , we obtain
  

 
 
x1
−2x2 − 2x4
−2
−2
x2  





x2
 =
 = x2  1  + x4  0  .
x3  

0
−1
−x4
x4
0
1
x4
So a basis for N (A) is
   
−2
−2 


   

1
 , 0  .
  −1

 0



0
1
(b) Find, explaining your reasoning, a basis for C(A), the column space of A.
A basis of C(A) is given by the columns of A corresponding to the pivot columns of rref(A). In
particular, the first and the third columns of A gives a basis for C(A):
   
1
1 


   

1
  , 2
1 4





1
3
(c) For the same matrix A consider the equation Ax = b where
 
b1
b2 

b=
b3 
b4
Math 51, Autumn 2015
Solutions to First Exam — October 13, 2015
Page 3 of 14
is a vector in R4 . Describe conditions on the components of b that are necessary and sufficient
for there to be a solution x to Ax = b. Your answer should be one or more equations involving
b1 , b2 , b3 , b4 .
Now find the

1 2
 1 2

 1 2
1 2
rref for the augmented matrix:


b1
1 3 b1
1 2 1 3


2 4 b2 
0 0 1 1 b2 − b1
→
 0 0 3 3 b3 − b1
4 6 b3 
3 5 b4
0 0 2 2 b4 − b1


1

 0
→

 0
0
2
0
0
0
0
1
0
0

2b1 − b2
2

1
b2 − b1

0 2b1 − 3b2 + b3 
0 b1 − 2b2 + b4
A necessary and sufficient condition for there to be a solution to Ax = b is that there is no
inconsistency in the rref:
(
2b1 − 3b2 + b3 = 0
b1 − 2b2 + b4
=0
Math 51, Autumn 2015
Solutions to First Exam — October 13, 2015
Page 4 of 14


5
3. (10 points) (a) Find the projection of the vector −4 onto the x-axis. Show your work.
1
 
5

The projection of −4 onto the x-axis is just the vector in the x direction which goes the “same
1
 
 
5
5
distance as −4” in the x direction. More formally, let v = −4, and let w be the projection
1
1
 
1
of v onto the x-axis. Then, w is defined to be the multiple of 0 such that v − w is orthogonal
0
 
 
1
5
to 0. Thus, we can easily see by inspection that w = 0.
0
0
Althoughthe
 answer is geometrically clear in this case, it also works to use the projection formula.
1
Let x = 0 (we may also replace x with any vector spanning the x-axis without changing our
0
calculation). Then, the problem asks for
   
5
1
5    
x·v
x= · 0 = 0 .
Projx (v) =
kxk2
1
0
0
(b) What is the shortest distance between the point (5, −4, 1) and the x-axis? Show your work.
The projection of a vector v onto a line can also be regarded as the closest point
 on that line to
5
v. Applying this principle together with the previous part, the closest point to −4 should be
1
 
5
0. Then, the distance is
0
   2  2
0 5
5 p
√
−4 − 0 = −4 = 02 + (−4)2 + 12 = 17.
1 1
0 To convince yourself of this geometrically, you may sketch these vectors, and you will see how
to calculate the distance essentially by using the Pythagorean theorem.
(c) Consider the line L parametrized by
{(5, −4, 1) + t(1, −1, 0) | t ∈ R}
in R3 . Find the length of the shortest line segment joining a point on L and a point on the x-axis.
Show your work.
(Hints: Such a line segment must be perpendicular to each line. Alternatively, find an expression
for the distance between each point of L and the x-axis in terms of t, and find its smallest value
over all values of t.)
Math 51, Autumn 2015
Solutions to First Exam — October 13, 2015
Page 5 of 14
There are several ways to do this problem.


5+t
Solution 1. Let v(t) = −4 − t, so that the points on L are exactly those which take the form
1
v(t) for some real number
 t. If we are given t, then we may calculate the projection of v(t) onto
5+t
the x-axis to be  0 , using the same procedure as in part (a). Then, by the same procedure
0
as in part (b), the distance from v(t) to the x-axis is
p
(−4 − t)2 + 12 .
(1)
Thus, to calculate the minimum distance, we simply need to find the value of t which minimizes
(1). This can be done by calculus (taking a derivative, solving for its zeroes, etc.). However,
in this case it is also easy to see that the expression (−4 − t)2 is always non-negative and is
minimized when t = −4. Thus, the minimum distance is 1.
Solution 2. Suppose that a and b are points on L and the x-axis which achieve the minimum
possible distance between the two lines. We may write

 

s
5+t



b = 0 ,
a = −4 − t ,
0
1
for some real numbers s and t. In order for a and b to achieve the minimal distance, it must be
the case that a − b is orthogonal to both L and the x-axis. Thus, we obtain the equations
 
 
1
1
(a − b) · 0 = 0.
(a − b) · −1 = 0,
0
0
When expanded out in terms of S and t, this gives
(5 + t − s) − (−4 − t) = 0,
5 + t − s = 0.
Solving for s and t, we find that we must have t = −4 and s = 1. Thus,
 
 
1
1
b = 0 ,
a = 0 ,
1
0
and so the distance from a to b is 1.
Solution 3. We can directly computea vector
v that is orthogonal to both L and the x-axis.

v1
Indeed, we must solve for coordinates v2  such that
v3
   
v1
1
v2  · −1 = 0,
0
v3
   
v1
1
v2  · 0 = 0.
0
v3
Solving this system, we end up with a solution
   
v1
0
v = v2  = 0 .
v3
1
Math 51, Autumn 2015
Solutions to First Exam — October 13, 2015
Page 6 of 14
(Of course, any multiple thereof will also work.)
Now, imagine two planes P1 and P2 , both orthogonal to v, with P1 containing L and P2 containing
the x-axis. (The fact that v is orthogonal to both L and the x-axis makes it possible to find
such planes.) Then, the distance between L and the x-axis will be the distance between P1 and
P2 . (If you cannot convince yourself of this, try to draw a picture. If you still can’t, maybe you
should focus on the other solutions.)
Now, how do we compute the distance between P1 and P2 ? Well, the shortest path from P1 to
P2 is by going along the direction of v. Next, we can pick any two points a ∈ P1 and b ∈ P2 . It
is convenient to pick a to be a point on L, so let’s pick
 
 
5
0
a = −4 ,
b = 0 .
1
0
Then, P2 is a translate of P1 by b − a, and so the distance between them is
 
5
Projv (b − a) = Projv −4 = 1.
1
Note: In this case, P1 and P2 were particularly simple; they were parallel to the xy plane. Thus,
some of the steps above could have been done by inspection, but we have given an approach that
would have worked in general.
Math 51, Autumn 2015
4. (10 points)
Solutions to First Exam — October 13, 2015
Page 7 of 14
Suppose that a is an unspecified real number. Consider the linear system:


 
1 2 3
1
1 1 1 x = 1
2 2 a
1
(a) Find, with justification, all values of a for which the above system has exactly one solution x.
(b) Find, with justification, all values of a for which the above system has no solution x.
(c) Find, with justification, all values of a for which the above system has an infinite number of
solutions x.
The linear system shown (of three equations
matrix

1
1
2
in three variables) is represented by the augmented

2 3 1
1 1 1 .
2 a 1
We row-reduce the matrix as follows:








1 2 3 1
1 2
3
1
1 2
3
1
1 2
3
1
1 1 1 1 → 0 −1 −2
0  → 0 1
0  → 0 1
0 .
2
2
2 2 a 1
0 −2 a − 6 −1
0 −2 a − 6 −1
0 0 a − 2 −1
At this point we can see that if a = 2, the last equation
(a − 2)z = −1
is inconsistent, so the system has no solutions.
If a 6= 2, then we can divide the last row by a − 2 to get


1 2 3 1
0 1 2 0  .
−1
0 0 1 a−2
At this point the three pivot 1’s have already appeared. We will be able to row-reduce the matrix
with no further complications. There are three pivots and no free variables, so there is a unique
solution.
So, for all a 6= 2, the equation has exactly one solution.
Since all possible real a are accounted for, there are no values of a for which the equation has infinitely
many solutions.
Math 51, Autumn 2015
Solutions to First Exam — October 13, 2015
Page 8 of 14
5. (8 points) The solutions to a system of linear equations can be interpreted as the intersection of a set
of (affine) subspaces of Rn . Match each of the systems of equations to the corresponding picture of an
(affine) subspace arrangement. (The axes might be rotated.) No justification required.
Answer by writing the number (no ) of the geometric pictureto the right of the matrix.
−1 1 0
5
2
4
−1 1 0
0
−1 1 0
1
0
0
 −1 −2 0 
−3 −1 0
−1 1 −2
0 1 2 0
no
no
no
2
1 0
0 −1 −5 −5
0 1
5 −5

15
no

0 1 2 4
 0 −1 2 4 
0 2 0 0

11
no

0 1 2 0
 0 −1 2 9 
0 2 0 0
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
no

12
no

1 1 0 0
 0 1 1 0 
0 0 1 0
13
The question was graded out of 8, with one point per correctly labelled system of equations.
A first observation: A system of m equations in 2 variables represents m lines in R2 (with the
possibility that some lines coincide). A system of m equations in 3 variables represents m planes in
R3 (again with the possibility that some planes coincide). This means we can immediately rule out
the graphics 7, 8, and 9 as possible answers for any system, and narrow down the options for each
remaining system of equations.
10
no
Math 51, Autumn 2015
Solutions to First Exam — October 13, 2015
Page 9 of 14
The solution to the system – the set of points that simultaneously satisfy each equation – is the
intersection of these lines or planes.
One way to approach these problems is to determine
the individual
lines or planes represented by each
−1
1
−2
row of each matrix (for example,
the
row
represents
the equation −x + y = 2, which
is the line y = x + 2. The row 0 −1 2 4 represents the equation −y + 2z = 4, which represents
a plane with normal vector [0, −1, 2]T in R3 . The following solutions will not consider specific lines
and planes, however, and instead look at qualitative features of the system, such as whether any lines
or planes are parallel, and the dimension of the space of solutions.
−1 1 0
represents two lines in R2 . Since the system is homogeneous,
The system of equations
−3 −1 0
the solution space must be a subspace of R2 . Since the rows [−1 1] and [−3 − 1] are (by inspection)
linearly independent, these lines are not parallel, and they must intersect in a single point, the origin.
The answer is 5.
−1 1 0
The system of equations
also corresponds to two lines in R2 , but since the normal
−1 1 −2
vectors [−1 1] and [−1 1] are collinear, these lines are parallel. Since the equations do not encode
the same line, these lines are parallel and distinct; the system has no solutions, and the lines do not
intersect. The answer is 2.


−1 1 0
The system  −1 −2 0  encodes three lines in R2 . None of the rows are collinear, so no two
2
1 0
lines are parallel. The system is homogeneous so all three lines must pass through the origin. The
answer is 4.
1 0 0 0
The system
encodes two planes in R3 . Since (by inspection) the normal vectors
0 1 2 0
[1, 0, 0]T and [0, 1, 2]T are not collinear, these planes are not parallel. The answer is 13.
0 −1 −5 −5
also encodes two planes in R3 . Their normal vectors [0, −1, −5]T
The system
0 1
5 −5
and [0, 1, 5]T are collinear, so these planes are parallel. The equations do not define the same plane
(the equation −y − 5z = −5 is not a scalar multiple of the equation y + 5z = −5), so the planes are
parallel and distinct. Alternatively, we can row-reduce the matrix to see that the system of equations
has no solutions, so the planes cannot intersect. The answer is 15.


0 1 2 4
The system  0 −1 2 4  gives three planes in R3 . Since no row is a scalar multiple of any
0 2 0 0
other row, no two planes coincide. Since no two normal vectors are collinear, no two planes are
parallel. By row-reducing the matrix, we see that the system is consistent and has a 1-parameter
space of solutions, so the planes must intersect in a line. The answer is 11. We can also complete this
problem by noticing that the column space is 2-dimensional (since at most two columns are linearly
independent), and the system is consistent since [4, 4, 0]T is in the column space – it is twice the third
column. This tells us that the system has solutions, and the rank-nullity theorem tells us that the
solutions are a line. Again we conclude that the answer is 11.


0 1 2 0
The system  0 −1 2 9  also describes three planes in R3 , where no two planes coincide and
0 2 0 0
no two planes are parallel. But this time, by row-reducing we find that the system is inconsistent, so
the planes have no common points of intersection. The answer is 12.


1 1 0 0
The system  0 1 1 0  encodes three distinct, non-parallel planes in R3 . We can row-reduce
0 0 1 0
Math 51, Autumn 2015
Solutions to First Exam — October 13, 2015
Page 10 of 14
to find that these planes intersect in a point. Or we can observe that the columns are linearly
independent, so the rank-nullity theorem implies that the nullspace (the intersection of the planes) is
{0}. The answer is 10.
Math 51, Autumn 2015
Solutions to First Exam — October 13, 2015
Page 11 of 14
6. (10 points)
(a) Complete the following sentence: A set of vectors {v1 , v2 , · · · , vk } is defined to be linearly dependent if
A set of vectors {v1 , · · · , vk } is linearly dependent if we can find scalars c1 , · · · , ck ∈ R not all
zero such that the linear combination c1 v1 + · · · + ck vk = 0 is the zero vector.
(b) Complete the following: A set V of vectors in Rn to be is defined to be a subspace if the following
three conditions hold
The three conditions that must be satisfied in order for V ⊂ Rn a subspace are that it contain
the zero vector, be closed under addition, and be closed under scalar multiplication. In other
words, we must have (i) 0 ∈ V , (ii) if v, w ∈ V , then v + w ∈ V , and (iii) if v ∈ V and c ∈ R is
any scalar, then cv ∈ V .
(c) Let A be a fixed n × n matrix. Define S to be the set of all vectors x ∈ Rn which satisfy:
Ax = −2x
Show that S is a subspace of Rn .
We need to show that the set of vectors x which satisfy Ax = −2x is a subspace. First, we
check that 0 is in this putative subspace. This claim is obvious: A0 = 0 = −20. Next, if
v and w satisfy the condition that Av = −2v, Aw = −2w, we can check that A(v + w) =
Av + Aw = −2v − 2w = −2(v + w), so v + w is also in the putative subspace. Nice. We’re
two-thirds of the way there. That’s so cool. Finally, if Av = −2v and c ∈ R, then we check
A(cv) = cAv = c(−2v) = −2cv = −2(cv) so cv is also in the set of vectors satisfying the given
condition. Hence, the set is in fact a subspace, as it satisfies the three properties listed in the
previous part.
Math 51, Autumn 2015
Solutions to First Exam — October 13, 2015
Page 12 of 14
7. (8 points)


7
(a) Find the coordinates (c1 , c2 ) of the vector u = −3 with respect to the basis {v1 , v2 } where
1
 
 
−1
2
v1 =  1  and v2 = 0.
1
2
In other words, express u as a linear combination u = c1 v1 + c2 v2 . Show your work.
Set up the system
 
   
−1
2
7





c1 1 + c2 0 = −3
1
2
1
which yields the linear system


−c1 + 2c2 = 7
c1
= −3


c1 + 2c2 = 1
The system is simple enough to solve it “manually” (without resorting to augmented matrices
and RREF): the second equation gives c1 = −3 and substituting into either the first or third
gives c2 = 2.
Thus
 
   
−1
2
7
−3  1  + 2 0 = −3
1
2
1
We can indeed verify that this works. The coordinates are (c1 , c2 ) = (−3, 2).
(b) Find the coordinates (c1 , c2 ) of u with respect to the basis elements v1 and v2 .
Justify your solution with a brief written explanation and/or a picture.
Hint: The solutions are integers.
The vectors appear to form a parallelogram with v1 the diagonal and u, v2 the sides. By the
Math 51, Autumn 2015
Solutions to First Exam — October 13, 2015
parallelogram rule, we conclude
v1 = u + v2
This means
u = v1 − v2 = 1v1 + (−1)v2
so the answer is
(c1 , c2 ) = (1, −1)
Page 13 of 14
Math 51, Autumn 2015
Solutions to First Exam — October 13, 2015
Page 14 of 14
8. (8 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.
(a) Given a matrix A, then N (A) = N (rref(A)).
T
F
SOMETIMES
Row operations applied to the augmented matrix do not change the set of solutions. So the
solutions to Ax = 0 and rref(A)x = 0 are the same.
T
F
SOMETIMES
(b) Given a matrix A, then dim(C(A)) = dim(C(rref(A))).
The set of columns of A corresponding to the pivots of rref(A) form a basis for C(A). The pivot
columns of rref(A) form a basis for C(rref(A)). The number of basis elements is the same.
(c) Given a vector b ∈ Rm and an m × m (square) matrix A for which T
F
SOMETIMES
the system Ax = b has exactly one solution, then the system Ax = 0
has infinitely many solutions.
If the system Ax = b has a solution xp then xp + xh , where xh ∈ N (A), is also a solution. So, if
there is a unique solution, N (A) = {0}. This is true for any shaped matrix A.
(d) Given a vector b ∈ Rm and an m × m (square) matrix A for which T
F
SOMETIMES
the system Ax = b has no solutions, then the system Ax = 0 has
infinitely many solutions.
If Ax = b has no solution, C(A) 6= Rm and the rank of A is < m. But rank + nullity = m so
nullity is > 0 and Ax = 0 has an infinite number of solutions. This uses the fact that A is square.
The answer would be“Maybe” in the general m × n case.
(e) Given an m × n matrix A with m < n, then dim(N (A)) > 0.
T
F
SOMETIMES
In this case there are more columns than rows, hence more than the number of pivots, so there is
at least one free variable.
(f) Given an m × n matrix A with m > n, then N (A) = {0}.
T
F
SOMETIMES
The conclusion N (A) = {0} holds if and only the columns are linearly independent. That may or
may not be the case.
(g) Given an m × n matrix A with m > n, then dim(C(A)) = m.
T
F
SOMETIMES
The dimension of C(A) equals the number of linearly independent columns of A. Since there are
n of them and n < m there can never be m linearly independent ones.
(h) Given an m×n matrix A and a set {v1 , . . . , vk } ⊂ Rn that is linearly T
SOMETIMES
F
independent, then the set {Av1 , . . . , Avk } is linearly independent.
This depends on the matrix A and the linearly independent set. For example, if the linearly
independent set equals {e1 , . . . , en } then the set {Ae1 , . . . , Aen } is the set of columns of A which
are linearly independent if and only if N (A) = {0}.