MATH 221 - 102
EXAM #1
Name:
Student ID:
Exam rules:
• No calculators, open books or notes are allowed.
• There are 5 problems in this exam. All problems are worth the same number
of marks.
• Use opposite empty pages if needed.
1
Problem 1. Find the general solution to the traffic flow problem as shown in the
picture. Write the general solution in the parametric vector form.
100
X1
200
X2
X3
X4
X5
300
Problem 2. Consider vectors
 
1
0

~u = 
 2 ,
−1


3
−2

~v = 
 1 ,
−2
 
b1
b2 
~b =   .
b3 
b4
(a)Find the equations that b1 , b2 , b3 , b4 must satisfy so that the three vectors are
linearly dependent.
(b)Find a vector ~b with b1 = 0, b2 = 2 and a nontrivial relation among ~u, ~v , ~b.
Problem 3. Recall that if masses mi are placed at points ~vi , then the center of
mass of this system is
1
v=
(m1~v1 + . . . + mn~vn ).
m1 + . . . + mn
Let
· ¸
· ¸
· ¸
· ¸
1
4
5
1
~v1 =
, ~v2 =
, ~v3 =
, ~v4 =
.
1
2
3
3
Assume that a total mass of 30g is distributed among the four points in such a way
that the center of mass is
· ¸
3
v=
.
2
(a)Find all possible distributions of masses m1 , m2 , m3 , m4 at the four points.
(Hint: one equation should come from the condition that the total mass is
30g.)
(b)Assuming that all masses are non-negative, find the solution m1 , m2 , m3 , m4
with m4 maximal possible.
Problem 4. Consider the system of

1 1
1 −1
0 2
equations in matrix form
    
2
x1
1




a · x2 = b 
−b
x3
0
where a and b are some numbers.
(a)Find all values of a and b such that the system is consistent.
(b)Find a, b and two distinct vectors ~u and ~v such that both ~u and ~v are solutions
to the matrix equation with parameters a and b. In the answer ~u and ~v should
be two vectors with concrete numbers, involving no variables.
Problem 5. Answer each question ”True” or ”False”. No explanation is necessary.
(a)If the augmented matrix [A|~b] of the system A~x = ~b is of size 4 × 5 and has 4
pivots, then there exists at most one solution to the equation A~x = ~b.
(b)There exist an infinite number of distinct 5 × 4 matrices in reduced row echelon
form having exactly 4 pivots.
(c)It is possible that A~x = ~b and A~x = ~c have the same solution vector ~x even
when ~b 6= ~c.
(d)Let ~vi be the vector in Rn with i ones, followed by n − i zeros:
 
1
 .. 
.
 
1
vi =  
0
.
 .. 
0
Then ~v1 , . . . , ~vn span Rn .
(e)If A~x = ~0 has only the trivial solution, then A~x = ~b has a unique solution for
any ~b.
(f)If A is a 5 × 5 matrix and for some ~b the system A~x = ~b has infinitely many
solutions, then the columns of A do not span R5 .