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Unformatted text preview: es that the ﬁrst n odd numbers add to n2 .
"
#" #
y1
D B1
y1 D B1
1
00
B1
D B2
1
10
B2
3 y1 C y2
gives y2 D B1 CB2
D
y1 C y2 C y3 D B3
y3 D
B2 CB3
0
11
B3
"
#
"
#
100
100
The inverse of S D 1 1 0 is A D 1 1 0 : independent columns in A and S !
111
011 4 The combination 0w1 C 0w2 C 0w3 always gives the zero vector, but this problem looks for other zero combinations (then the vectors are dependent, they lie in a plane):
w2 D .w1 C w3 /=2 so one combination that gives zero is 1 w1 w2 C 1 w3 :
2
2 5 The rows of the 3 by 3 matrix in Problem 4 must also be dependent: r 2 D 1
.r 1
2 The column and row combinations that produce 0 are the same: this is unusual.
"
#
135
1 2 4 has column 3 D 2 .column 1/ C column 2
6 cD3
113
"
#
101
1 1 0 has column 3 D
cD 1
column 1 C column 2
011
"
#
000
2 1 5 has column 3 D 3 .column 1/ column 2
cD0
336 C r 3 /. 7 All three rows are perpendicular to the solution x (the three equations r 1 x D 0 and r 2 x D 0 and r 3 x D 0 tell us this). Then the whole plane of the rows is perpendicular
to x (the plane is also perpendicular to all multiples c x ).
2
32 3
x1 0 D b1
x1 D b1
1000
b1
x2 x1 D b2
x2 D b1 C b2
6 1 1 0 0 7 6 b2 7
8
D4
D A 1b
x3 x2 D b3
x3 D b1 C b2 C b3
1 1 1 0 5 4 b3 5
x4 x3 D b4
x4 D b1 C b2 C b3 C b4
1111
b4 9 The cyclic difference matrix C has a line of solutions (in 4 dimensions) to C x D 0: 2 1
61
40
0 0
1
1
0 0
0
1
1 32 3 2 3
23
1
x1
0
c
0 7 6 x2 7 6 0 7
6c 7
D
when x D 4 5 D any constant vector.
0 5 4 x3 5 4 0 5
c
1
x4
0
c Solutions to Exercises
z2 z1 D b1
10 z3 z2 D b2
0 z3 D b3 7
z1 D
z2 D
z3 D b1 b2
b2 b3
b3
b3 D " 1
0
0 1
1
0 1
1
1 #" b1
b2
b3 # D 1 b 11 The forward differences of the squares are .t C 1/2 t 2 D t 2 C 2t C 1 t 2 D 2t C 1.
Differences of the nth power are .t C 1/n t n D t n t n C nt n 1 C . The leading
term is the derivative nt n 1 . The binomial theorem gives all the terms of .t C 1/n .
12 Centered difference matrices of even size seem to be invertible. Look at eqns. 1 and 4:
2
32 3 2 3
2
32
3
0
1
00
x1
b1
First
x1
b2 b4
0
1 0 7 6 x2 7 6 b2 7 solve
61
6 x2 7 6 b1
7
D
D
40
5
1
0 1 5 4 x3 5 4 b3 5 x2 D b1 4 x3 5 4 b4
0
0
10
x4
b4
x3 D b4
x4
b1 C b3
13 Odd size: The ﬁve centered difference equations lead to b1 C b3 C b5 D 0. x2
x3
x4
x5 x1
x2
x3
x4 D b1
D b2
D b3
D b4
D b5 Add equations 1; 3; 5
The left side of the sum is zero
The right side is b1 C b3 C b5
There cannot be a solution unless b1 C b3 C b5 D 0. 14 An example is .a; b/ D .3; 6/ and .c; d / D .1; 2/. The ratios a=c and b=d are equal. Then ad D bc . Then (when you divide by bd ) the ratios a=b and c=d are equal! Problem Set 2.1, page 40
1 The columns are i D .1; 0; 0/ and j D .0; 1; 0/ and k D .0; 0; 1/ and b D .2; 3; 4/ D 2i C 3j C 4k. 2 The planes are the same: 2x D 4 is x D 2, 3y D 9 is y D 3, and 4z D 16 is z D 4. The
3
4
5 6
7 8 9 solution is the same point X D x . The columns are changed; but same combination.
The solution is not changed! The second plane and row 2 of the matrix and all columns
of the matrix (vectors in the colu...
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.
 Spring '12
 Minki
 Mass

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