aidT) anoniqo 5. A = -STT XIMEN OSS A = -1 15. LNGA batt 1 0 0 0 0 0 0 T ₁. vino gniau o bouber wor -3dw to 7000 7 b= 3-5 -6 4 8 1 0 7700 00 ON-L oqque ES ensupse Do A

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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5

hisa Isnogcib sei d
wod2.isnogsib
oldim
1. A =
A =
Siga eriti
2. A =
A =
2.5 EXERCISES
vities
In Exercises 1-6, solve the equation Ax = b by using the LU
factorization given for A. In Exercises 1 and 2, also solve Ax = b
by ordinary row reduction. $1 ei tuqni ar
HER
-3 5
42
obsl
100
-1 1 0
owl smid
2 -5 1
1
-1
3 -7 -2
1812151
1
own gat
6 -4
7130
2
2
3. A = -6
8
gies C
4 3 -5
-4 -5 7
8
6-8
].
0
1
0
-1
0
0
-1
qqu
ill:
18
2017 ind
, b =
5
2
b= =
i][
0
L
3-71-27
0 -2 -1
UE 2nist
0 -
4
Find an LU factorization of A =
PRACTICE PROBLEM
0-2
0
has only three pivot columns, so the method of Example 2 will produce only the first
three columns of L. The remaining two columns of L come from 15.]
-4
6
27
3 -5
2
2
2
-2 ,b= 0
5
0
wo
od2
TO PRACTICE'R
and world 80
donnos on
231 odi
istenen szorlw
olugmo) ce
(15) d
LI to
2-4-2 3
6-9 -5 8
2 -7 -3 9
4 -2 -2 -1
-6 3 3 4
Aankope
of bodromid,
madinggis diqqA) sai
A =
-3
4 -1
4
(Paloviqtoris
LOA
A =
5. A =
YAM
ed as A VW
airT)
A =
2000 SAJTAM, ald
esiuns 0132d-2
4
4. A = 1-31 1, b =
=
33007 5
0
1 0
I
1
A gnizo,TI
[Note: It will turn out that A
1
bet indien UST
0 0 0
1/2 1
3/2
2
16
0
-5 1
mo (or
8
2 -1 27
Alo 199
-3 4
0
of bon
0
01
100
210
-1 0
1
-4 3 -5
1
2 -7 -7 -6
-1 26
4
8
-4 -1 9
01
-Abai7 81
0715wolsed kl 01
-52 Ianogsib and no
7 pH) aslugnain
ainsinsoslag worvino
-2
snrw mizo,alA
4
ogneria
U=AMI .00
bsouber wor
ban A
quod
moitos2 to
-2-1
0-6
0 0
-2 -4 -3 adw to los
191
7
0
3
On ai sevib
0 1
0
0
0
0
0
,b=
A paoqqu2.15
sanaupos
Do A
-2 -4 -3
-3 1
0 2
0
0
0
1
1
Transcribed Image Text:hisa Isnogcib sei d wod2.isnogsib oldim 1. A = A = Siga eriti 2. A = A = 2.5 EXERCISES vities In Exercises 1-6, solve the equation Ax = b by using the LU factorization given for A. In Exercises 1 and 2, also solve Ax = b by ordinary row reduction. $1 ei tuqni ar HER -3 5 42 obsl 100 -1 1 0 owl smid 2 -5 1 1 -1 3 -7 -2 1812151 1 own gat 6 -4 7130 2 2 3. A = -6 8 gies C 4 3 -5 -4 -5 7 8 6-8 ]. 0 1 0 -1 0 0 -1 qqu ill: 18 2017 ind , b = 5 2 b= = i][ 0 L 3-71-27 0 -2 -1 UE 2nist 0 - 4 Find an LU factorization of A = PRACTICE PROBLEM 0-2 0 has only three pivot columns, so the method of Example 2 will produce only the first three columns of L. The remaining two columns of L come from 15.] -4 6 27 3 -5 2 2 2 -2 ,b= 0 5 0 wo od2 TO PRACTICE'R and world 80 donnos on 231 odi istenen szorlw olugmo) ce (15) d LI to 2-4-2 3 6-9 -5 8 2 -7 -3 9 4 -2 -2 -1 -6 3 3 4 Aankope of bodromid, madinggis diqqA) sai A = -3 4 -1 4 (Paloviqtoris LOA A = 5. A = YAM ed as A VW airT) A = 2000 SAJTAM, ald esiuns 0132d-2 4 4. A = 1-31 1, b = = 33007 5 0 1 0 I 1 A gnizo,TI [Note: It will turn out that A 1 bet indien UST 0 0 0 1/2 1 3/2 2 16 0 -5 1 mo (or 8 2 -1 27 Alo 199 -3 4 0 of bon 0 01 100 210 -1 0 1 -4 3 -5 1 2 -7 -7 -6 -1 26 4 8 -4 -1 9 01 -Abai7 81 0715wolsed kl 01 -52 Ianogsib and no 7 pH) aslugnain ainsinsoslag worvino -2 snrw mizo,alA 4 ogneria U=AMI .00 bsouber wor ban A quod moitos2 to -2-1 0-6 0 0 -2 -4 -3 adw to los 191 7 0 3 On ai sevib 0 1 0 0 0 0 0 ,b= A paoqqu2.15 sanaupos Do A -2 -4 -3 -3 1 0 2 0 0 0 1 1
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