18.06 Problem Set 5 Solutions
Problem 1. If a parabola t the data exactly we would get a solution (v1 , v2 , v3 ) to the system
2
1 1 1
5
1 2 4 v1
1 3 9 v2 = 7 .
v3
1
1 4 16
Since this system
18.06 Problem Set #6 Solutions
1.
a) If we denote the number of subscribers to Star at k-th year by sk , and the number of
subscribers to Times by tk , then the conditions in the problems tell us sk+1
Course 18.06, Fall 2002: Quiz 1, Solutions
1 (a) For example
4
2
2 , z = 6
w=
4
0
or
3
5
4 , z = 3u v = 0
w =u +v =
2
6
(b), (c) For example
2 1
M = 1 3
2 0
x3 free variable. Let x3 = 1 then
0
Course 18.06, Fall 2002: Quiz 2, Solutions
1 (a) The columns of A are linearly dependent, but the column space is spanned by A =
3
.
1
Use this matrix in the formula for the projection matrix:
PC = A(
8
8S
8
8
8 8
8
8
8
iX
iX
i
iX
q
Q
8
i
X
f
d b
d r
8 cfw_v S
v
q
Q
8
8
f
8
d
v d T
cfw_v y
i egw
i
X 8
R D T R F T r r T R UC D Yu F R rC D H T R H U R
@xIV'GVedXVQesEG
18.06 Problem Set #3 Solutions
1. The set in (a) cant be a basis because two vectors can span an at most 2 dimensional vector space,
while R3 is 3 dimensional. The sets in (b) and (d) cant be bases si
18.06 Exam 3 Solutions
1. a) M = 0.7 0.5 0.3 0.5 .
b) It is the eigenvector for M corresponding to eigenvalue 1, (0.5, 0.3). c) After many years, the percentage of people drinking two kinds of coffee
9
9 l l
l
9 rq F
l 9 9
l
9 l l
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AjA|qruxkrauWrTTu`2qrTf|aesAyXaksU
Ur U b` x d h` dr `r y c h
asayxfqraau`hx aA|yWeajAiRsA8A33rb yW|m am fD
W
18.06
Professor A.J. de Jong
Your name is:
Please circle your recitation:
Exam 2
April 9, 2003
Important: Briey explain all of your answers.
1 (29 pts.)
(a) Compute the determinant of the following ma
18.06 Problem Set #1 Solutions
1a) (2 points) In order for the two vectors to be parallel we must have (a, 4) = c(2, 5) for some
constant c. Thus, we need a = 2c and 4 = c5. It follows that c = 4 and
18.06 Exam 1 #1 Solutions
1 a)
v x = 0 x1 + 2x2 + x3 = 0
w x = 0 2x1 + 4x2 + 3x3 = 0
So the set to be found is the nullspace of the matrix A =
1 2 1
. The row echelon form
2 4 3
1 2 1
. The second var
18.06
Fall 2003
Quiz 1
October 1, 2003
Your name is:
Please circle your recitation:
1. M2 S. Harvey
7. T11 N. Ganter
2. M2 D. Ingerman
8. T12 N. Ganter
3. M3 S. Harvey
9. T12 S. Francisco
4. T10 B. Su
18.06
Fall 2004
Quiz 2
November 15, 2004
Your name is:
Please circle your recitation:
1. M2 A. Brooke-Taylor
7. T11 V. Angeltveit
2. M2 F. Liu
8. T12 V. Angeltveit
3. M3 A. Brooke-Taylor
9. T12 F. Roc
18.06 Exam 2 #1 Solutions
1. The row echelon form of
1 2 1 4
A is 0 1 2 3 . So we nd that a basis for R(AT ) is
0 0 0 0
and a basis for N (A) is cfw_(3, 2, 1, 0), (2, 3, 0, 1). Similarly, row
0 1
1 2
18.06
Fall 2004
Quiz 1
October 13, 2004
Your name is:
Please circle your recitation:
1. M2 A. Brooke-Taylor
7. T11 V. Angeltveit
2. M2 F. Liu
8. T12 V. Angeltveit
3. M3 A. Brooke-Taylor
9. T12 F. Roch