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Unformatted text preview: rs shows a; b; c; d; e; f are unchanged in the product
(important for A D LU in Section 2.6).
MM 1 D .In U V / .In C U.Im V U / 1 V / .this is testing formula 3/
D In U V C U.Im V U / 1 V U V U.Im V U / 1 V .keep simplifying/
D In U V C U.Im V U /.Im V U / 1 V D In .formulas 1; 2; 4 are similar/
43 4 by 4 still with T11 D 1 has pivots 1; 1; 1; 1; reversing to T D UL makes T44 D 1. 44 Add the equations C x D b to ﬁnd 0 D b1 C b2 C b3 C b4 . Same for F x D b. CA 1 B (and d c b=a is the correct
second pivot of an ordinary 2 by 2 matrix).
The example problem has
1
10
4
56
SD
.
33 D
4
01
65
2 45 The block pivots are A and S D D Solutions to Exercises 21 1
A 1 D A 1 .I C
AB/ 1 . So I CBA and I CAB are both invertible or both singular when A is invertible.
(This remains true also when A is singular : Problem 6.6.19 will show that AB and BA
have the same nonzero eigenvalues, and we are looking here at D 1.) 46 Inverting the identity A.I C BA/ D .I C AB/A gives .I C BA/ Problem Set 2.6, page 102
10
10 x
5
1 `21 D 1 multiplied row 1; L D
times
D
D c is Ax D b :
11
11 y
2
11 x
5
D
.
12 y
7
1 0 c1
5
5
2 Lc D b is
D
, solved by c D
as elimination goes forward.
1 1 c2
7
2
11 x
5
3
U x D c is
D
, solved by x D
in back substitution.
01 y
2
2 3 `31 D 1 and `32 D 2 (and `33 D 1): reverse steps to get Au D b from U x D c :
4 5 6 7 8 9 1 times .x Cy Cz D 5/C2 times .y C2z D 2/C1 times .z D 2/ gives x C3y C6z D 11.
"
#" # " #
"
#" # " #
"#
1
5
5
111
5
5
2D
7 ; Ux D
12
x D 2 ; xD
2.
Lc D 1 1
121
2
11
1
2
2
"
#"
#"
#
1
210
210
01
0 4 2 D 0 4 2 D U . With E 1 as L, A D LU D
EA D
301
635
005
"
#
1
01
U.
301
#"
#
"
#
"
#
"
1
111
100
1
01
21
A D 0 2 3 D U . Then A D 2 1 0 U is
021
001
006
021
the same as E211 E321 U D LU . The multipliers `21 ; `32 D 2 fall into place in L.
#"
#"
#
#"
"
100
1
1
1
21
2 2 2 . This is
1
1
E32 E31 E21 A D
345
21
3
1
1
#
"
#
"
100
101
0 2 0 D U . Put those multipliers 2; 3; 2 into L. Then A D 2 1 0 U D LU .
002
321
#
#"
#"
"
#"
1
1
1
1
a1
a
1
1
1
.
D
E D E32 E31 E21 D
ac b
c1
b
1
1
c1
The multipliers are just a; b; c and the upper triangular U is I . In this case A D L and
its inverse is that matrix E D L 1 .
#
#"
"
#"
1
d e g d D 1; e D 1, then l D 1
110
f h f D 0 is not allowed
2 by 2: d D 0 not allowed; 1 1 2 D l 1
i
no pivot in row 2
mn1
121 Solutions to Exercises 22 10 c D 2 leads to zero in the second pivot position: exchange rows and not singular.
11 12 13 14 15 16 c D 1 leads to zero in the third pivot position. In this case the matrix is singular.
"
#
"
#
248
2
3
A D 0 3 9 has L D I (A is already upper triangular) and D D
I
007
7
"
#
124
A D LU has U D A; A D LDU has U D D 1 A D 0 1 3 with 1’s on the
001
diagonal.
24
10 24
10 20 12
AD
D
D
D LDU ; U is LT
4 11
21 03
21 03 01
"
#"
#"
#"
#"
#
1
14...
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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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