Geometrically

Pat Rossi
MTH 3331 Practice Test #2 - Answers
Summer 2013
Name
1. ~
(a) u + v
b + d
b
(a, b)
u
v
(a + c, b + d)
(0, 0)
a
a + c
(b) u
v
b
d
(0, 0)
(a, b)
u
a
v
u ­ v
c
(c, d)
2. jjujj = 13; jjvjj = p 61
3. Cauchy-Schwarz Inequality: If u and v are vectors in < n ; then ju vj jjujj jjvjj :
Equality holds exactly when u =kv:
4. Triangle Inequality: If u and v are vectors in < n ; then jju + vjj jjujj + jjvjj :
Equality holds exactly when u =kv: and give its geometric interpretation. Geometrically,
this means that no one side of a triangle has length greater than the sum of the
lengths of the other two sides.
b + d
b
(a, b)
u
v
(a + c, b + d)
(0, 0)
a
a + c
5. u v = 15

6. ~
(a) jjujj = 1
(b) u is a unit vector.
7. cos () = 0
8. cos () = 2
7 p = p 10
10 35
2 2 3
9.
4
3
x 1
x 2
5 =
x 3
4
5
1
0
2
5 + k 4
2
1
1
3
5
10. Assume that each of the following matrices is the matrix of coe¢ cients of a homogeneous
system of equations. Decide whether has any solution other than the trivial
solution. If so, give the solution.
2
3
1 0 0
(a) The matrix 4 0 1 1 5 is row equivalent to the identity matrix and therefore
0 0 1
has only the trivial solution. Alternately, the matrix has rank 3 and therefore has
(b)
(c)
(d)
only the trivial solution.
2 3 2 3
x 1
1
4 x 2
5 = k 4 1 5
x 3 1
2 3 2 3
x 1
2
6 x 2
7
4 x 3
5 = k 6 1
7
1 4 0 5 + k 2
x 4 0
2
4
3
x 1
x 2
x 3
2
5 = k 4
1
0
0
11. Ax = 0 has only the trivial solution, x = 0:
2 3 2 3 2 3
1
17
x 1
2
2
12. 4 x 2
5 = 4 1 5 + k 4 7 5
x 3 0 1
13. See solutions.
3
5
2
6
4
0
0
1
1
3
7
5
14. See solutions.
2 3 2
x 1
x 2
15. The
x 3
6 x 4
=
7 6
4 x 5
5 4
x 6
6
0
1
0
2
0
3 2
+ k 1 7 6
5 4
2
1
0
0
0
0
3 2
+ k 2 7 6
5 4
1
0
4
1
0
0
3 2
+ k 3 7 6
5 4
1
0
2
0
2
1
3
7
5
2

16. x 1 + y 1 is also a particular solution of the system Ax = b:
17.
x
y
1 = 2k
4 = 2k
18. x + 3y + 4z = 16
19. 2x + 2y 3z = 1
x 2 = k
20. y 1 = 3k
z 2 = 3k
2
3 12
3
5
21. A 1 = 4 1 5 2 5
2 8 3
2 3
x
2 3
17
22. 4 y 5 = 4
z
8 5
11
23. 2x + 3y + 4z = 9
24. 4x 4y + 3z = 10
25. ~
(a)
(b)
(c)
2
4
2
6
4
2
6
4
3
x 1
x 2
5 =
x 3
3
x 1
x 2
7
x 3
x 4
3
x 1
x 2
x 3
7
x 4
x 5
2
4
2
5 = 6
4
2
3
0
5
0
3
1
2
5 = k 1 6
4
3
2
5 + k 4
3
2
7
5 + k 6
2
1
0
0
0
3
4
1
2
1
3
5
3
2
1
7
0 5
0
2
7
5 + k 2 6
4
4
0
2
3
1
3
7
5
3