Unformatted text preview: ns are
and
,AD4
and U D 4
.
17
26
0 0 1 15
0
0 1 15
0101
0
000
Elimination leaves the diagonal matrix diag.3; 2; 1/ in 3x D 3; 2y D 2; z D 4. Then
x D 1; y D 1; z D 4.
A.2; W/ D A.2; W/ 3 A.1; W/ subtracts 3 times row 1 from row 2.
The average pivots for rand(3) without row exchanges were 1 ; 5; 10 in one experiment—
2
but pivots 2 and 3 can be arbitrarily large. Their averages are actually inﬁnite ! With
row exchanges in MATLAB’s lu code, the averages :75 and :50 and :365 are much
more stable (and should be predictable, also for randn with normal instead of uniform
probability distribution).
If A.5; 5/ is 7 not 11, then the last pivot will be 0 not 4.
Row j of U is a combination of rows 1; : : : ; j of A. If Ax D 0 then U x D 0 (not true
if b replaces 0). U is the diagonal of A when A is lower triangular.
The question deals with 100 equations Ax D 0 when A is singular.
(a) Some linear combination of the 100 rows is the row of 100 zeros.
(b) Some linear combination of the 100 columns is the column of zeros.
(c) A very singular matrix has all ones: A D eye(100). A better example has 99
random rows (or the numbers 1i ; : : : ; 100i in those rows). The 100th row could
be the sum of the ﬁrst 99 rows (or any other combination of those rows with no
zeros).
(d) The row picture has 100 planes meeting along a common line through 0. The
column picture has 100 vectors all in the same 99dimensional hyperplane. Problem Set 2.3, page 63
" #
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#"
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#
100
100
100
010
010
5 1 0 ; E32 D 0 1 0 ; P D 0 0 1
100 D 001.
1 E21 D
001
071
010
001
100
2 E32 E21 b D .1; 5; 35/ but E21 E32 b D .1; 5; 0/. When E32 comes ﬁrst, row 3
feels no effect from row 1.
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#"
#"
#
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100
100
1
0
0
1
0
0
410;010;0
1
0 M D E32 E31 E21 D
4
1
0:
3
001
201
0
2
1
10
2
1 Solutions to Exercises 12
23
23
2
1
1
6 7 E21 6 7 E31 6
4 Elimination on column 4: b D 4 0 5 ! 4 4 5 ! 4
0
0
original Ax D b has become U x D c D .1; 4 ; 10/. Then
1
z D 5; y D 2 ; x D 1 : This solves Ax D .1; 0; 0/.
2 1 3 2 1 3 7 E32 6 7
4 5 ! 4 4 5. The
2
10
back substitution gives 5 Changing a33 from 7 to 11 will change the third pivot from 5 to 9. Changing a33 from 7 to 2 will change the pivot from 5 to no pivot.
2
32 3
23
237
1
4
6
76 7
67
6 Example: 4 2 3 7 5 4 3 5 D 4 4 5. If all columns are multiples of column 1,
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there is no second pivot. 1 4 7 To reverse E31 , add 7 times row 1 to row 3. The inverse of the elimination matrix
3
2
3
2 8 9 10 11 12 100
100
7
6
7
6
1
E D 4 0 1 0 5 is E D 4 0 1 0 5.
701
701
ab
a
b
MD
and M * D
. det M * D a.d `b/ b .c `a/
cd
c `a d `b
reduces to ad b c !
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#
100
M D 0 0 1 . After the exchange, we need E31 (not E21 ) to act on the new row 3.
110
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#"
#
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#
101
101
201
E13 D 0 1 0 I 0 1 0 I E31 E13 D 0 1 0 : Test on the identity matrix!
001
101
101
#
"
122
An example with two negative pivots is A D 1 1 2 . The diagonal entries can
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change sign during elimination.
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rows and
1
2
3
987
also columns T...
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 Spring '12
 Minki
 Linear Algebra, Matrices, Dot Product, Mass, Diagonal matrix, Row

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