Download Exam 2 Sol

APPM 2360: Section exam 2
7:00pm – 8:30pm, October 24, 2012.
ON THE FRONT OF YOUR BLUEBOOK write: (1) your name, (2) your student ID number,
(3) recitation section (4) your instructor’s name, and (5) a grading table. Text books, class notes,
and calculators are NOT permitted. A one-page crib sheet is allowed.
Problem 1: (20 points) Consider the following system of equations
ẋ = 3x + 3y + 3
ẏ = 4x − 2y + 4
(a) (4 points) Determine and draw the nullclines for this system.
(b) (4 points) Determine and draw the directions along each nullcline.
(c) (4 points) Determine and sketch the direction in the various quadrants (i.e. between the
nullclines)
(d) (4 points) What are the equilibria for this system? And of what type are they?
(e) (4 points) Draw some possible solution curves.
solution
(a) v-nullcline: ẋ = 0 = 3x + 3y + 3 ⇒ y = −x − 1
h-nullcline: ẏ = 0 = −2y + 4x + 4 ⇒ y = 2x + 2
see figure
(b) v-nullcline: ẋ = 3( 12 y − 1) + 3y + 3 = 29 y which is positive for y > 0 and negative for y < 0
h-nullcline: ẏ = −2(−x − 1) + 4x + 4 = 6x + 6 = 6(x + 1) which is positive for x > −1 and
negative for x < −1
see figure “Nullclines”
(c) see figure “Solution”
(d) The equilibria are at the intersection of a h-nullcline and a v-nullcline:
−x − 1 = 2x + 2 ⇒ 3x = −3 ⇒ x = −1.
And from the v-nullcline (or h-nullcline): y = 1 − 1 = 0. The only equilibrium is at (−1, 0)
and it is unstable.
(e) see figure “Solution”
Problem 2: (20 points) Give a brief answer to each question. Show all work.
1 2
−1 2
, and let B =
, compute C = (A + B)2 − A2 + B 2 .
3 1
1 1




1
0 −1
5 4 2
1 , and let B =  1 3 1 , verify that B−2I = (A−1 )T .
(b) (4 pts) Let A =  2 −2
−2
1
1
2 3 4
Hint: You don’t have to compute an inverse.
(a) (3 pts) Let A =
(c) (3 pts) Let A and B be 3 × 3 matrices with det(A) = −5 and det(B) = 2 and C an
invertible 3 × 3 matrix, find the values of
(i) det AC (BC)−1
(ii) det B−1 A
T (iii) det AT B A−1
(d) (3 pts) Show that the set of all real symmetric 2 × 2 matrices, S2×2 , is a vector space. Hint:
Note that the set of all real 2 × 2 matrices, M2×2 , is a vector space.
(e) (4 pts) Let V be the plane in R3 defined by the equation
x1 + x2 + x3 = 0
T
Show that ~v1 = [1 0 − 1]T and ~v
2 = [0 1 − 1]
form a basis for V.
−1
2
1 −3
(f) (3 pts) Given A =
and B =
, find a 2 × 2 matrix Y such that
3 −7
1
3
AY = B.
Solutions
6 4
(a) C = (A + B) −
+
= AB + BA =
2 10



1
2 −2
3 4 2
1   1 1 1  = I = (B − 2I)AT , so (B − 2I) = (AT )−1 =
(b) AT (B − 2I) =  0 −2
−1
1
1
2 3 2
−1
T
(A ) .
(c) (i) det AC (BC)−1 = det(A) det(C) det(C−1 ) det(B−1 ) = det(A)/ det(B) = −5/2
2
A2
B2
(ii) det B−1 A = det(B−1 ) det(A) = det(A)/ det(B) = −5/2
T (iii) det AT B A−1
= det(AT ) det(B) det(A−1 ) = det(A) det(B)/ det A = det B = 2
(d) From the hint, S2×2 ∈ M2×2 and we only need to test for closure. For any scalars α, β ∈ R
we have
a1 a2
b1 b2
αa1 + βb1 αa2 + βb2
α
+β
=
∈ S2×2
a2 a3
b2 b3
αa2 + βb2 αa3 + βb3
and so we have closure under matrix addition and scalar multiplication of matrices, thus
S2×2 is a vector space.
(e) 
Let ~v 
be a general vector
 in the plane:



x1
x1
1
0
 x2  = 
 = x1  0  + x2  1  x1 and x2 being arbitrary the vector
x2
x3
−x1 − x2
−1
−1
~v1 and ~v2 span the plane.




1
0
1 0
1  with RREF  0 1  and both columns are
The matrix of the problem is  0
−1 −1
0 0
pivot columns. Thus ~v1 and~
v
are
linearly
independent.
{~
v
v2 } is a basis of the plane.
1, ~
2
−7 −2
1 −3
−9 15
(f) Y = A−1 B =
=
−3 −1
1 3
−4 6
Problem 3: (20 points) Consider the matrix


2 1 0 2
 2 1 1 2 

A = [~v1 ~v2 ~v3 ~v4 ] = 
 2 0 0 1 .
2 2 0 2
(a) (8 points) Compute the determinant of the matrix A.
(b) (8 points) Is A invertible? If yes, find the inverse. If not, justify your answer.
(c) (4 points) ~v1 , ~v2 , ~v3 , ~v4 are the column vectors of the matrix A. For what values of a,b,c,d
is ~v5 = [a b c d]T in Span{~v1 , ~v2 , ~v3 , ~v4 }? Justify your answer.
solution:
2 1 2 1 2 2 1 − 1
(a) det(A) = −1 2 0 1 = −1 −2 2 2 = −1 (−2(2 − 4) − 1(4 − 2)) =
2 2 2 2 2 −2.
(b) A is invertible since det(A) 6= 0.




1
2 1 0 2 1 0 0 0
1 0 0 0 −1 0
1
2
 2 1 1 2 0 1 0 0 
 0 1 0 0 −1 0
0
1 




 2 0 0 1 0 0 1 0  →  0 0 1 0 −1 1

0
0
2 2 0 2 0 0 0 1
0 0 0 1
2 0 −1 −1
(c) Since det(A) 6= 0 there is a solution to Ax = ~v5 for any a, b, c, d ∈ R.
PLEASE TURN OVER
Problem 4: (20 points) Consider a linear system whose augmented matrix is of the form


2 −1 −1 0
 1
1
a 0 
−3
0
2 0
(a) (7 points)
(b) (3 points)
(c) (3 points)
(d) (7 points)
Solutions
For what values of a will the system have a unique solution?
What is the unique solution to the system given your answer in (a)?
For what values of a will the system be inconsistent?
Suppose a = −1, solve the system.
(a) By the method of cofactors, the determinant of the matrix A in the system Ax = 0 is 3(1 + a).
Therefore the system has a unique solution provided a 6= −1.
(b) If Det(A) 6= 0 the unique solution must be x = 0.
(c) No values of a are inconsistent, the system is homogeneous.
(d)






1
1 −1 0
1
1 −1 0
2 −1 −1 0
 1
1 −1 0  →  2 −1 −1 0  →  0 −3
1 0 →
−3
0
2 0
−3
0
2 0
0
3 −1 0




1 1 −1 0
1 0 − 23 0
 0 1 −1 0  →  0 1 −1 0 
3
3
0 0
0 0
0 0
0 0
The third column is a non-pivot column, therefore set z = t where t ∈ R to obtain
 2 
3


x = t


1
3





1
(Problem 5: (20 points) Let PE2 define the vector space of all polynomials of degree ≤ 2. All
vector elements v ∈ PE2 therefore have the form
e = a0 + a1 t + a2 t2 .
v
(a) (4 points) What is the dimension of PE2 ?
(b) (10 points) Do the vectors
v1 = 1 + t + t 2 ,
v2 = −1 + t2 ,
v3 = −2 + t + 4t2
form a basis for PE2 ? No credit will be given for a simple yes or no answer. Show your
work.
(c) (6 points) Given v = t2 , is v ∈ Span{v1 , v2 , v3 }?
solution:
(a) dim=3
(b) GIven the equivalence to R3 : v1 = (1, 1, 1)T , v2 = (−1, 0, 1)T , v3 = (−2, 1, 4)T . The
polynomials are linearly dependent:

 
 
 

1 −1 −2 0
1 −1 −2 0
1 −1 −2 0
1 0 1 0
 1 0
1 0 ∼ 0 1
3 0 ∼ 0 1
3 0 ∼ 0 1 3 0 
1 1
4 0
0 2
6 0
0 0
0 0
0 0 0 0
Not needed but
 

1 −1 −2 a0
1 −1 −2
 1 0
1 a1  ∼  0 1
3
1 1
4 a2
0 2
6

1 0
∼ 0 1
0 0
Span[v1 , v2 , v3 ] is a2 − 2a1 + a0 = 0.
(c) No v = t2 ∈
/ Span{v1 , v2 , v3 }

 
1 −1 −2 0
1 −1 −2
 1 0
1 0 ∼ 0 1
3
1 1
4 1
0 2
6
 

a0
a0
1 −1 −2

a1 − a0  ∼  0 1
3
a1 − a0
a2 − a0
0 0
0 a2 − 2a1 + a0

1
a1

a1 − a0
3
0 a2 − 2a1 + a0
 
 

1 −1 −2 0
1 0 1 0
0
0 ∼ 0 1
3 0 ∼ 0 1 3 0 
1
0 0
0 1
0 0 0 1