(20) 1.A-01-2 22-1 A=λI = 20 011 2 22-1- def (A-X)| = 1-1/12-12-22-24+022 1-) ((1-1)(-1-\)-2(-21)-2(01-1-2-21-27) + (001-211-22) = 1-X (843) - 2 (4 to =-83+12-3x-5=0. (X+1) (x²-2x+5)=0. Eigen Values λ=-1, = 1+2, λ=1-2 Eigen vectors λ=-1: Av Av 1120\X 01-2014 = = शि 122-1112 X+24=-x, 2y= -2x = -x 4-22=-9 2x+2y-Z = -Z -X-22=x, -22=2x12=-X. Let x=1, then y=-1, Z=-1 Then eigen vector V₁ = (1). A = 1+29, Av= Av: [120\[X] x+2y= (1+2i/x Y-27=(HZily 2x+2y-2=11+2:12 Tet U₂ = ($) + = []). 2 1X-2Z = (1+2;)(ix), iX-27=x-28 22= -2x, Let x=1, then y=i, Z=1. Z-X λ= 1-21, Av=λv= x+2y = (1-201x, 8y = 11-2;1X x 19 = ix 9-27=11-2dy (-ix)-22 = (1-2 -ix. Z=X Let x=1, then y = -1, Z=1 Let V3 = 0 4 0 abc Then p= (708). (84) Pr1 = della 189 (a; -9c) (ah-96) Tefl 96 det (P) = -1/30 | -0/00 | +1/03 = -1 -1/-3)-0(0)+1(0-1-31) -1/-3)-0(01 + 1(0-1-31) = 3-0+3=6 10 10 11 P-1=17 11 01 Foil Fiil 1101 10 --1) (1-11-10 + - + 100 1 0+2 1+1 6 31-11-0)-0 (α(-1)-(-11(01) +1(0(01-(-1) (3))
(20) 1.A-01-2 22-1 A=λI = 20 011 2 22-1- def (A-X)| = 1-1/12-12-22-24+022 1-) ((1-1)(-1-\)-2(-21)-2(01-1-2-21-27) + (001-211-22) = 1-X (843) - 2 (4 to =-83+12-3x-5=0. (X+1) (x²-2x+5)=0. Eigen Values λ=-1, = 1+2, λ=1-2 Eigen vectors λ=-1: Av Av 1120\X 01-2014 = = शि 122-1112 X+24=-x, 2y= -2x = -x 4-22=-9 2x+2y-Z = -Z -X-22=x, -22=2x12=-X. Let x=1, then y=-1, Z=-1 Then eigen vector V₁ = (1). A = 1+29, Av= Av: [120\[X] x+2y= (1+2i/x Y-27=(HZily 2x+2y-2=11+2:12 Tet U₂ = ($) + = []). 2 1X-2Z = (1+2;)(ix), iX-27=x-28 22= -2x, Let x=1, then y=i, Z=1. Z-X λ= 1-21, Av=λv= x+2y = (1-201x, 8y = 11-2;1X x 19 = ix 9-27=11-2dy (-ix)-22 = (1-2 -ix. Z=X Let x=1, then y = -1, Z=1 Let V3 = 0 4 0 abc Then p= (708). (84) Pr1 = della 189 (a; -9c) (ah-96) Tefl 96 det (P) = -1/30 | -0/00 | +1/03 = -1 -1/-3)-0(0)+1(0-1-31) -1/-3)-0(01 + 1(0-1-31) = 3-0+3=6 10 10 11 P-1=17 11 01 Foil Fiil 1101 10 --1) (1-11-10 + - + 100 1 0+2 1+1 6 31-11-0)-0 (α(-1)-(-11(01) +1(0(01-(-1) (3))
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Find D and e^(tA) = P e^(tD) P^(-1).
![(20)
1.A-01-2
22-1
A=λI =
20
011 2
22-1-
def (A-X)| = 1-1/12-12-22-24+022
1-) ((1-1)(-1-\)-2(-21)-2(01-1-2-21-27) + (001-211-22)
= 1-X (843) - 2 (4 to
=-83+12-3x-5=0.
(X+1) (x²-2x+5)=0.
Eigen Values λ=-1, = 1+2, λ=1-2
Eigen vectors
λ=-1: Av Av
1120\X
01-2014 = = शि
122-1112
X+24=-x, 2y= -2x = -x
4-22=-9
2x+2y-Z = -Z
-X-22=x, -22=2x12=-X.
Let x=1, then y=-1, Z=-1
Then eigen vector V₁ = (1).
A = 1+29, Av= Av:
[120\[X]
x+2y= (1+2i/x
Y-27=(HZily
2x+2y-2=11+2:12
Tet U₂ = ($) + = []).
2
1X-2Z = (1+2;)(ix), iX-27=x-28 22= -2x,
Let x=1, then y=i, Z=1.
Z-X](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe742bcfc-6571-4eb6-8020-0d244ddd1492%2F2ec5fbd9-c08f-40de-810f-90a960cddbae%2Fmstsyr9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(20)
1.A-01-2
22-1
A=λI =
20
011 2
22-1-
def (A-X)| = 1-1/12-12-22-24+022
1-) ((1-1)(-1-\)-2(-21)-2(01-1-2-21-27) + (001-211-22)
= 1-X (843) - 2 (4 to
=-83+12-3x-5=0.
(X+1) (x²-2x+5)=0.
Eigen Values λ=-1, = 1+2, λ=1-2
Eigen vectors
λ=-1: Av Av
1120\X
01-2014 = = शि
122-1112
X+24=-x, 2y= -2x = -x
4-22=-9
2x+2y-Z = -Z
-X-22=x, -22=2x12=-X.
Let x=1, then y=-1, Z=-1
Then eigen vector V₁ = (1).
A = 1+29, Av= Av:
[120\[X]
x+2y= (1+2i/x
Y-27=(HZily
2x+2y-2=11+2:12
Tet U₂ = ($) + = []).
2
1X-2Z = (1+2;)(ix), iX-27=x-28 22= -2x,
Let x=1, then y=i, Z=1.
Z-X
Transcribed Image Text:λ= 1-21, Av=λv=
x+2y = (1-201x, 8y = 11-2;1X x 19 = ix
9-27=11-2dy
(-ix)-22 = (1-2 -ix. Z=X
Let x=1, then y = -1, Z=1
Let V3 =
0
4
0
abc
Then p= (708). (84)
Pr1 = della
189 (a; -9c) (ah-96)
Tefl
96
det (P) = -1/30 | -0/00 | +1/03 = -1
-1/-3)-0(0)+1(0-1-31)
-1/-3)-0(01 + 1(0-1-31) = 3-0+3=6
10
10
11
P-1=17
11
01
Foil Fiil 1101
10
--1) (1-11-10
+
-
+
100 1
0+2
1+1 6
31-11-0)-0 (α(-1)-(-11(01) +1(0(01-(-1) (3))
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