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 is over determined, we seek the vector x = (B, C, D) wh
18.06 Problem Set #6 Solutions
1.
a) If we denote the number of subscribers to Star at kth year by sk , and the number of
subscribers to Times by tk , then the conditions in the problems tell us sk+1 = .6sk +.1tk ,
tk+1 = .4sk + .9tk . So the Markov matr
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(AT A)1 AT =
3
1
1
10
3 1
=
1
10
9 3
3 1
1
(b) As befo
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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 since these sets cannot be linearly
independent. (If the
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 will converge to one that is proportional to the steady
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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 matrix
0 1 0 0 0
1 1 1 1
1 1 1 1
0 1 1 0
0 1 1 0
1 1 1 1
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 a = 8 .
5
5
1b) (2 points) We look for those values of
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 variable, x2 , is free and the vector (2, 1, 0) is a basis
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. Sutton
10. T1 K. Cheung
5. T10 C. Taylor
11. T1 B. Tenner
18.06
Fall 2004
Quiz 2
November 15, 2004
Your name is:
Please circle your recitation:
1. M2 A. BrookeTaylor
7. T11 V. Angeltveit
2. M2 F. Liu
8. T12 V. Angeltveit
3. M3 A. BrookeTaylor
9. T12 F. Rochon
4. T10 K. Cheung
10. T1 L. Williams
5. T10 Y. Rubin
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
. So a basis for C(A) is cfw_(1, 0, 1), (0, 1, 2) and
18.06
Fall 2004
Quiz 1
October 13, 2004
Your name is:
Please circle your recitation:
1. M2 A. BrookeTaylor
7. T11 V. Angeltveit
2. M2 F. Liu
8. T12 V. Angeltveit
3. M3 A. BrookeTaylor
9. T12 F. Rochon
4. T10 K. Cheung
10. T1 L. Williams
5. T10 Y. Rubins