Unformatted text preview: ) Any solution can be xp
p
1
6
33 x
2
is shorter (length 2) than
D
. Then
(c)
(length 2)
6
1
33 y
0
(d) The only “homogeneous” solution in the nullspace is x n D 0 when A is invertible. Solutions to Exercises 35 14 If column 5 has no pivot, x5 is a free variable. The zero vector is not the only solution
15 16
17 18
19
20 21 22 23 24 25
26 to Ax D 0. If this system Ax D b has a solution, it has inﬁnitely many solutions.
If row 3 of U has no pivot, that is a zero row. U x D c is only solvable provided
c 3 D 0. Ax D b might not be solvable, because U may have other zero rows needing
more ci D 0.
The largest rank is 3. Then there is a pivot in every row. The solution always exists.
The column space is R3 . An example is A D Œ I F for any 3 by 2 matrix F .
The largest rank of a 6 by 4 matrix is 4. Then there is a pivot in every column. The
solution is unique. The nullspace contains only the zero vector. An example is A D
R D Œ I F for any 4 by 2 matrix F .
Rank D 2; rank D 3 unless q D 2 (then rank D 2). Transpose has the same rank!
Both matrices A have rank 2. Always AT A and AAT have the same rank as A.
"
#"
#
100
10
1
0
10 3
410
02
2
3.
A D LU D
I A D LU 2 1 0
21 0
301
031
0 0 11
5
"# "#
"#
"#
"# "#
"#
x
4
1
1
x
4
1
1 Cz
0
0 . The second
(a) y D 0 C y
(b) y D 0 C z
z
0
0
1
z
0
1
equation in part (b) removed one special solution.
If Ax 1 D b and also Ax 2 D b then we can add x 1 x 2 to any solution of Ax D B :
the solution x is not unique. But there will be no solution to Ax D B if B is not in
the column space.
For A; q D 3 gives rank 1, every other q gives rank 2. For B; q D 6 gives rank 1, every
other q gives rank 2. These matrices cannot have rank 3.
11
b1
x1
1
Œx D
has 0 or 1 solutions, depending on b (b)
D
(a)
1
b2
x2
Œ b has inﬁnitely many solutions for every b (c) There are 0 or 1 solutions when A
has rank r < m and r < n: the simplest example is a zero matrix. (d) one solution
for all b when A is square and invertible (like A D I ).
(a) r < m, always r n (b) r D m, r < n (c) r < m; r D n (d) r D m D n.
#
"
#
"
#
"
10
2
244
244
0 3 6 !RD 0 1
2 and 0 3 6 ! R D I .
005
000
00
0 27 If U has n pivots, then R has n pivots equal to 1. Zeros above and below those pivots make R D I .
" #
2
120
1235
1230
1200
1;
!
!
; xn D
28
001
0048
0040
0010
0
Free x2 D 0 gives x p D . 1; 0; 2/ because the pivot columns contain I .
"
#
"#
"
1000
0
10
29 Œ R d D 0 0 1 0 leads to x n D 1 ; Œ R d D 0 0
0000
0
00
no solution because of the 3rd equation
1
.
2
0
1
0 #
1
2:
5 Solutions to Exercises 36 23
23
#
4
2
4
6 37
6 07
3 ; 4 5; x n D x3 4 5.
30
0
1
2
2
0
"
#
"#
11
1
0
31 For A D 0 2 , the only solution to Ax D 2 is x D
. B cannot exist since
1
03
3
2 equations in 3 unknowns cannot have a unique solution.
2
3
2
32
3
131
1
1
31
1 27
61 2 37
61 1
7 60
32 A D 4
factors into LU D 4
and the rank
5 40
2 4 65
221
0 05
115
1201...
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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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