Solutions to Math 51 Second Exam — May 12, 2016
1. (10 points) Suppose S : R2 → R2 and T : R2 → R2 are the following linear transformations:
• S rotates a vector v by 60" degrees
counterclockwise and then multiplies the x-coordinate by 2.
#
1
1
(For example, S
= √3 .)
0
2
• T takes each vector v to −v and then multiplies the x-coordinate by
1
.
2
(a) Write down the matrix of S.
The matrix for counter clockwise rotation by 60 degrees is given by
√ #
" 1
3
−
cos 60◦ − sin 60◦
2
2
√
.
=
3
1
sin 60◦ cos 60◦
2
2
The matrix for the transformation that multiplies the x-coordinate of a vector by 2 is given by
2 0
.
0 1
Hence the matrix of S is given by
√
"
2 0 √21
3
0 1
3
2
−
1
2
2
#
"
=
1
√
√ #
− 3
3
2
1
2
.
(b) Let M denote the matrix of T ◦ S. Find M .
0
1
be the standard basis vectors in R2 . By definition of T we have
and e2 =
Let e1 =
1
0
− 21
T (e1 ) =
;
0
0
T (e2 ) =
.
−1
Hence the matrix of T is given by
− 21
0
0
.
−1
Using the matrix of S obtained in part (a) we obtain
√ # "
1
"
1 − 3
−2 0
−√12
√
=
M=
3
1
0 −1
− 23
2
2
√
3
2
− 12
#
.
(c) Find M 6 ; simplify your answer as much as possible. (Hint: interpret the linear transformation
T ◦ S geometrically; what happens if you keep applying it successively 2, 3, 4, etc., times?)
Observe that T ◦ S is a rotation by 240 degrees counterclockwise. This can verified geometrically
by checking T ◦ S rotates both e1 and e2 counterclockwise by 240 degrees or simply by observing
cos 240◦ − sin 240◦
M=
sin 240◦ cos 240◦
It follows that (T ◦ S), applied 6 times in succession, rotates every point in the plane by 1440
(= 360 × 4) degrees counterclockwise, which amounts to four full rotations. Hence T ◦ S, applied
Math 51, Spring 2016
Solutions to Second Exam — May 12, 2016
six times in succession, is the identity transformation whose matrix M 6 is given by
1 0
6
M =
.
0 1
Page 2 of 14
Math 51, Spring 2016
Solutions to Second Exam — May 12, 2016
Page 3 of 14
 
2
A 0 = e1 ,
0
 
 
3
2
2. (10 points) Let A be a 3 × 3 matrix satisfying:
A 2 = e1 + e2 , A 0 = e1 + e3 .
0
3
h1i
(Recall that e1 , e2 , e3 are the standard basis vectors in R3 ; that is, e1 = 0 , and so on.)
0
A−1 .
(a) Find
(Hint: you do not need to compute A to do this; suppose v2 is the second column
−1
of A . What do you know about Av2 ?)
Do note that A is indeed invertible. This is so because its column space must contain e1 , e1 + e2 ,
e1 + e3 , and span(e1 , e1 + e2 , e1 + e3 ) = R3 , so A has full rank: 3.
 
 
 
2
2
2
Now, A 0 = e1 implies A−1 e1 = 0. Hence the first column of A−1 is 0.
0
0
0
 
 
3
3
Similarly A 2 = e1 + e2 implies A−1 e1 + A−1 e2 = 2, so
0
0
 
     
3
3
2
1
A−1 e2 = 2 − A−1 e1 = 2 − 0 = 2 .
0
0
0
0
Thus, the second column of A−1
 
1
is 2.
0
 
 
2
2
Finally A 0 = e1 + e3 implies A−1 e1 + A−1 e3 = 0, so
3
3
     
 
0
2
2
2
−1
−1







A e3 = 0 − A e1 = 0 − 0 = 0 .
3
0
3
3
Therefore the third column of A−1
 
0

is 0. All in all we obtain
3

A−1
(b) Solve for x in the following equation:
We have
(c) Find det(A).

2 1 0
=  0 2 0 .
0 0 3
 
7

Ax = 4 .
2
  
   
7
2 1 0
7
18
−1  





4 = 0 2 0
4 = 8 .
x=A
2
0 0 3
2
6
Math 51, Spring 2016
Solutions to Second Exam — May 12, 2016
Note that A−1 is a diagonal matrix and det(A−1 ) = 2 · 2 · 3 = 12. Hence
det(A) =
1
1
= .
det(A−1 )
12
Page 4 of 14
Math 51, Spring 2016
Solutions to Second Exam — May 12, 2016
Page 5 of 14
3. (10 points) Let C be the curve in R2 defined by x2 + 4xy + 5y 2 = 2.
(a) Find the equation of the line tangent to C at (−1, 1).
Let f (x, y) = x2 + 4xy + 5y 2 . Then C is the level set given by f (x, y) = 2. We compute the
partial derivatives of f :
∂f
= 2x + 4y
∂x
so
∂f
= 4x + 10y,
∂y
∂f
(−1, 1) = 2 · (−1) + 4 · 1 = 2
∂x
∂f
(−1, 1) = 4 · (−1) + 10 · 1 = 6,
∂y
The equation for the tangent line to C at (−1, 1) is
∂f
∂f
(−1, 1)(x + 1) +
(−1, 1)(y − 1) = 0,
∂x
∂y
i.e.
2(x + 1) + 6(y − 1) = 0.
1 4
. Show that
(b) Let T :
→
be the linear transformation whose matrix is given by A =
2 3
the image of the curve C under T is the circle x2 + y 2 = 10.
R2
R2
Note that the matrix A is invertible, so every point in R2 can be written in the form
x + 4y
x
x
1 4
=
A
=
2x + 3y
y
2 3
y
x + 4y
in exactly one way. Now we see that a point
2x + 3y
equation x2 + y 2 = 10 precisely when
of this form lies on the circle with
(x + 4y)2 + (2x + 3y)2 = 10.
This equation is the same as
x2 + 8xy + 16y 2 + 4x2 + 12xy + 9y 2 = 10,
i.e.
5x2 + 20xy + 25y 2 = 10
and dividing by 5 yields
x2 + 4xy + 5y 2 = 2.
This is precisely the equation of the curve
shown that
C. Hence have
a point is on the given
x
x + 4y
x
circle if and only if it is of the form A
=
for a point
on the curve C. This
y
2x + 3y
y
means that the circle with equation x2 + y 2 = 10 is the image of the curve C under the linear
transformation T .
(c) The curve C is an ellipse in R2 (this follows from part (b), but you may assume this without
proof). Find the area enclosed by C. (Hint: use determinants and the result of part (b).)
Since the circle x2 + y 2 = 10 is the image of the curve C under the linear transformation given
Math 51, Spring 2016
Solutions to Second Exam — May 12, 2016
Page 6 of 14
by the matrix A, we know
(Area enclosed by circle x2 + y 2 = 10) = | det(A)| · (Area enclosed by ellipse C).
√
√ 2
The circle x2 + y 2 = 10 has radius 10 and area π 10 = 10π. The matrix A has determinant
det(A) = 1 · 3 − 4 · 2 = 3 − 8 = −5. Thus, | det(A)| = 5. Plugging these values in, we get
10π = 5 · (Area enclosed by ellipse C),
so the area enclosed by the ellipse C is 2π.
Math 51, Spring 2016
Solutions to Second Exam — May 12, 2016
Page 7 of 14
4. (10 points) Below is a collection of level sets of a function f : R2 → R at whole-number values. You
may assume that f and its first and second derivatives are continuous, and the length scales in the
x- and y-directions are equal.
Math 51, Spring 2016
Solutions to Second Exam — May 12, 2016
Page 8 of 14
Please refer to the level-set diagram of the function f : R2 → R, as well as the stated assumptions
on f , as given on the facing page. For the questions below, you do not need to justify your answers.
(a) Sketch, on the plot, the direction of the gradient vector ∇f at B.
At B, the gradient vector points downwards.
(b) Sketch, on the plot, the direction of the gradient vector ∇f at F.
At F, the gradient vector points upwards and slightly to the left.
Note: In both (a) and (b) the gradient vector is in the direction of greatest increase, and
should be perpendicular to the nearest level sets.
(c) (Circle one)
(d) (Circle one)
(e) (Circle one)
(f) (Circle one)
∂f
at A is:
∂y
NEGATIVE
ZERO
1
Let v =
; then Dv f at C is:
1
1
1
; then Dv f at D is:
Let v = √2
1
1
1
Let v = √2
; then Dv f at E is:
1
√1
2
POSITIVE
NEGATIVE
ZERO
POSITIVE
NEGATIVE
ZERO
POSITIVE
NEGATIVE
ZERO
POSITIVE
Note: In each of (c), (d), (e), (f), start at the point in question, and take a small step in the
relevant direction. Does the function increase, decrease or stay the same?
∂ 2f
at E is:
NEGATIVE
POSITIVE
∂x2
∂ 2f
(h) (Circle one)
at E is:
NEGATIVE
POSITIVE
∂y 2
∂ 2f
(i) (Circle one)
at A is:
NEGATIVE
POSITIVE
∂x∂y
To examine a second derivative, look at the first derivative, take a small step in an appropriate
direction, and look at the first derivative again. How has it changed?
(g) (Circle one)
Math 51, Spring 2016
Solutions to Second Exam — May 12, 2016
Page 9 of 14
5. (8 points) Let G : R2 → R2 be defined by
G(x, y) = (x2 − y 2 − 4x + 4, 2xy − 4y).
(a) Compute DG(1, 0); show your steps.
The matrix of partial derivatives DG is
DG(x, y) =
2x − 4 −2y
.
2y
2x − 4
At the point (1, 0), this is
−2 0
DG(1, 0) =
.
0 −2
(b) Let H : R2 → R2 be defined by H(x, y) = G(G(G(x, y))). Show that H(1, 0) = (1, 0).
We calculate that
G(1, 0) = (1 − 0 − 4 + 4, 0 − 0) = (1, 0).
Thus G fixes the point (1, 0), so G applied three times to (1, 0) is (1, 0).
(c) Use linear approximation to estimate H(0.95, 0.03); simplify your answer as much as possible.
We calculate as follows, using the chain rule:
0.95 − 1
H(0.95, 0.03) ≈H(1, 0) + DH(1, 0)
0.03 − 0
1
−0.05
=
+ DG(G(G(1, 0)))DG(G(1, 0))DG(1, 0)
0
0.03
3 −0.05
−2 0
1
+
=
0.03
0 −2
0
−0.05
−8 0
1
+
=
0.03
0 −8
0
0.40
1
=
+
−0.24
0
1.4
=
.
−0.24
Math 51, Spring 2016
Solutions to Second Exam — May 12, 2016
Page 10 of 14
6. (10 points) Suppose that f : R2 → R is a differentiable function with the following properties:
• f (1, −2) = 5
• The derivative of f at the point a = (1, −2) is Df (1, −2) = −1 3 .
3 2
• The Hessian of f at the point a = (1, −2) is Hf (1, −2) =
.
2 1
(a) Give a unit vector u in R2 for which moving away from the point a = (1, −2) in the direction u
initially leads to approximately no change in the value of f .
We want to find a unit vector u ∈ R2 such that Du f = 0. Since Du f = ∇f · u, we want to find
a unit vector u such that ∇f · u = 0, i.e. such that u is orthogonal to ∇f . Now
−1
∇f = Df T =
,
3
−1
x
so we need to find a vector orthogonal to
. That is, we want to find a vector x =
such
3
y
that
x
−1
= −x + 3y = 0.
·
y
3
−1
2
, which we find by setting
This yields a line in R , so there are two directions orthogonal to
3
−3
3
. In order to make these unit vectors, we divide by their length, which
and
y = ±1:
−1
1
p
√
√
is 32 + 12 = (−3)2 + (−1)2 = 10. Thus, the two unit vectors in directions along which f
doesn’t change are
"
# "
#
u=
√3
10
√1
10
,
−3
√
10
−1
√
10
.
(b) Use linear approximation to estimate f (1.02, −2.03); show all your steps, and simplify your final
answer as much as possible.
We first compute the linear approximation L(x, y) to f at the point (1, −2). We have
x−1
x−1
L(x, y) = f (1, −2) + Df (1, −2)
= 5 + −1 3
= 5 − (x − 1) + 3(y + 2).
y+2
y+2
So we have
f (1.02, −2.03) ≈ L(1.02, −2.03) = 5 − (1.02 − 1) + 3(−2.03 + 2) = 4.89.
(c) Use a second-order Taylor polynomial (i.e., quadratic approximation) to estimate f (1.1, −1.9);
show all your steps, and simplify your final answer as much as possible.
We first compute the second-order Taylor polynomial p2 (x, y) of f at the point (1, −2). We have
1
x−1
x−1
x − 1 y + 2 Hf (1, −2)
p2 (x, y) = f (1, −2) + Df (1, −2)
+
y+2
y+2
2
x−1
3 2 x−1
1
x−1 y+2
= 5 + −1 3
+
y+2
2 1 y+2
2
3
1
= 5 − (x − 1) + 3(y + 2) + (x − 1)2 + 2(x − 1)(y + 2) + (y + 2)2 .
2
2
Math 51, Spring 2016
Solutions to Second Exam — May 12, 2016
Page 11 of 14
So we have
f (1.1, −1.9) ≈ p2 (1.1, −1.9)
3
1
= 5 − (1.1 − 1) + 3(−1.9 + 2) + (1.1 − 1)2 + 2(1.1 − 1)(−1.9 + 2) + (−1.9 + 2)2 = 5.24.
2
2
Math 51, Spring 2016
Solutions to Second Exam — May 12, 2016
Page 12 of 14
7. (10 points) Let g(x, y, z) = x2 + 4xy + 2y 2 − z 2 .
(a) Let S be the surface in R3 defined by g(x, y, z) = 6. Find the equation of the tangent plane to S
at the point (−1, 3, 1).


 
2x + 4y
10
∇g = 4x + 4y , therefore ∇g(−1, 3, 1) =  8 . The equation of the tangent plane is then
−2z
−2


  

x − (−1)
10
x+1
∇g(−1, 3, 1) ·  y − 3  = 0 ⇐⇒  8  · y − 3 = 0 ⇐⇒
z−1
−2
z−1
10(x + 1) + 8(y − 3) − 2(z − 1) = 0 ⇐⇒ 10x + 8y − 2z = 12
(b) Find a 3 × 3 symmetric matrix A for
which the quadratic form associated to A is given by g, i.e.,
hxi
for which g(x, y, z) = x y z A yz .


1 2 0
A = 2 2 0 
0 0 −1
(c) Is the matrix A (or equivalently the quadratic form g) positive definite, negative definite, indefinite,
or none of these? Justify your answer.
Notice that g(1, 0, 0) = 1 > 0 and g(0, 0, 1) = −1 < 0. Therefore A is indefinite.
Math 51, Spring 2016
Solutions to Second Exam — May 12, 2016
Page 13 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 two n × n matrices A, B, then AB = BA.
Example, where AB

1 0
 0 1
0 0
T
F
MAYBE
= BA holds:

 
 


0
2 0 0
2 0 0
2 0 0
1 0 0
0  0 3 0  =  0 3 0  =  0 3 0  0 1 0 
1
0 0 4
0 0 4
0 0 4
0 0 1
Example, where AB = BA does not hold:
1 0
0 1
0 1
=
,
0 −1
1 0
−1 0
but
0 1
1 0
1 0
0 −1
(b) Given a 5 × 4 matrix C and a 4 × 5 matrix D, then CD is invertible.
=
0 −1
1 0
T
F
.
MAYBE
Both the matrix C and the matrix D have rank at most 4. Therefore the rank of CD is at most
4 (because rank(CD) ≤ rank(C) and rank(CD) ≤ rank(D)). But CD is a 5 × 5-matrix. Hence
CD is not invertible.
(c) Given a 5 × 4 matrix C and a 4 × 5 matrix D, then DC is invertible.
T
F
MAYBE
F
MAYBE
Example, where DC is invertible:

1
 0

 0
0
0
1
0
0
0
0
1
0
0
0
0
1

 1
0 
 0
0 
 0
0 
 0
0
0
and the matrix on the RHS is invertible.
Example, where DC is not invertible:



1 0 0 0 0 
 0 0 0 0 0 


 0 0 0 0 0 

0 0 0 0 0
0
1
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0

0
0
0
0
0
0
0
0
0
0
0
0
0
0
0


1 0 0

  0 1 0
=
  0 0 1

0 0 0

0

  0
=
  0

0
0
0
0
0
0
0
0
0

0
0 

0 
1

0
0 

0 
0
and the matrix on the RHS is not invertible.
(d) Given two invertible n × n matrices A, B, then AB is invertible.
T
If A and B are invertible, then det(A) 6= 0 and det(B) 6= 0 and therefore det(AB) = det(A) det(B) 6=
0. Hence AB is invertible.
(e) Given an invertible 4 × 4 matrix A, then swapping the first two rows of A
produces a matrix B satisfying det(B) = det(A).
T
F
MAYBE
Swapping two rows multiplies the determinant by −1. Hence det(B) = − det(A). Since A is
invertible, we have det(A) 6= 0, so det(B) = − det(A) 6= det(A).
(f) Given an m×n matrix M and a vector b ∈ Rm , the function f : Rn → Rm
defined by f (x) = M x + b has derivative Df (a) = M at every point
a ∈ Rn .
T
F
MAYBE
Math 51, Spring 2016
Solutions to Second Exam — May 12, 2016
Page 14 of 14
See homework problem 17 on homework 4 (the constant term +b has no influence on the partial
derivatives).
(g) Given a function f : Rn → R with continuous second derivatives at
a ∈ Rn satisfying fxx (a) ≥ 0 and fyy (a) ≥ 0, then fxy (a) ≥ 0 as well.
T
F
MAYBE
Example, where fxy (a) ≥ 0: Consider f : R2 → R given by f (x, y) = x2 + xy + y 2 . Then for
every point a ∈ R2 we have fxx (a) = 2 ≥ 0 and fyy (a) = 2 ≥ 0. We also have fxy (a) = 1 ≥ 0
for every point a ∈ R2 .
Example, where fxy (a) < 0: Consider f : R2 → R given by f (x, y) = x2 − xy + y 2 . Then for
every point a ∈ R2 we have fxx (a) = 2 ≥ 0 and fyy (a) = 2 ≥ 0. But we have fxy (a) = −1 < 0
for every point a ∈ R2 .
(h) Given an n × n symmetric matrix A whose entries are all positive, then
A is positive definite.
Example, where A is positive definite: The matrix
2 1
1 2
T
F
MAYBE
has only positive entries and the corresponding quadratic form is
2x2 + 2xy + 2y 2 = x2 + y 2 + (x + y)2 .
This is positive definite.
Example, where A is not positive definite: The matrix
1 100
100 1
has only positive entries and the corresponding quadratic form is
Q(x, y) = x2 + 200xy + y 2
Since Q(1, 1) = 202 > 0 and Q(1, −1) = −198 < 0, this is indefinite. Hence A is indefinite and
therefore not positive definite.
Please do not detach this page. If you use any of this space to continue your answer, please clearly
indicate the problem number(s).