Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Question at end of answer of solution from a
![question 8.
The given matrix is
A² (603)
4
-2
-
A and for each
a) Here we have to find the eigen Valves of
eigen Valve we have to find a Corresponding eigen rector in the form of
, where a is a Positive number.
eigen Value of Athen
Let & be an
LA-X 1²0
6- X 4
ㅕ
-0
-2
-3)
=1 (6-X) (-3-X) + 8 = 0
=1-18-6 X + 3 + ³ + 8 = 0
-3 -10 = 0
-1
=1 X-SX 72 X-10=0
= > >=-5) +2(X - 5) = 0
= (x+2)(x-5) = 0
کر 2-= X
Take AF-2 and 1₂ s
let V=
Then (A-1₁I) V-6
7(A +2₁) V=0 [Xi = -2]
= 6 + ² 4
2
- 2
[V₂] be an eigen vector of A Corresponding to di= -2
- 3+2
V₁
√₂/
0
8 V₁T Y√₂ = 0 and 2V₁ V₂ =0
+
2V,
ond - 2V₁ + V₂ =6
ㅋ
= 2V₁ + V₂ =0
V₂=0
8
12
V₁ = 1 and V₂ = 2
>
as
on
We
Solving,
get
Thus we can take V₁ = (-2)
√₂](https://content.bartleby.com/qna-images/question/b4f3c4ec-fe5b-4d14-92c7-f03ecdbdce74/bfa5716e-f14b-4bc7-967b-7a9e6469005b/4n1r94a_processed.jpeg)
Transcribed Image Text:question 8.
The given matrix is
A² (603)
4
-2
-
A and for each
a) Here we have to find the eigen Valves of
eigen Valve we have to find a Corresponding eigen rector in the form of
, where a is a Positive number.
eigen Value of Athen
Let & be an
LA-X 1²0
6- X 4
ㅕ
-0
-2
-3)
=1 (6-X) (-3-X) + 8 = 0
=1-18-6 X + 3 + ³ + 8 = 0
-3 -10 = 0
-1
=1 X-SX 72 X-10=0
= > >=-5) +2(X - 5) = 0
= (x+2)(x-5) = 0
کر 2-= X
Take AF-2 and 1₂ s
let V=
Then (A-1₁I) V-6
7(A +2₁) V=0 [Xi = -2]
= 6 + ² 4
2
- 2
[V₂] be an eigen vector of A Corresponding to di= -2
- 3+2
V₁
√₂/
0
8 V₁T Y√₂ = 0 and 2V₁ V₂ =0
+
2V,
ond - 2V₁ + V₂ =6
ㅋ
= 2V₁ + V₂ =0
V₂=0
8
12
V₁ = 1 and V₂ = 2
>
as
on
We
Solving,
get
Thus we can take V₁ = (-2)
√₂
![148)
an eigen vector of a
Let
x = (2₂) bc
(A-XI)x=0
X 2 = S
to
There fore
Vector is
A
Corresponding to A X1 = -2
eigen
= (A -51) X = 0 [°° >₂=5]
(625 $35) (2) (8)
=1
-1 (1-2-3) (62) -(8
=1 x₁ +4x₂ = 0 and -22₁ -8 2₂=0
=1
IC₁ + 4x₂ = 0 and x₁ 4x2 = 0
=1
хсi + Чх 2 : 0
On solving, we get 20₁=4 and 2² = 1
Thus we
Can
take
V₂ = (-1) as
an
Vector of A Corresponding to
x = 6x + 4y
y = -2x - 3y
on
X₁ = -2 is an eigen Value of A
V2 >
answer above.
eigen Vector
25 Then
of Corresponding
(e)
Use your answer to part (a) to find the general solution of the following
system of linear differential equations, writing the solution as an
equation for x and an equation for y:
and Corresponding eigen](https://content.bartleby.com/qna-images/question/b4f3c4ec-fe5b-4d14-92c7-f03ecdbdce74/bfa5716e-f14b-4bc7-967b-7a9e6469005b/jseyutd_processed.jpeg)
Transcribed Image Text:148)
an eigen vector of a
Let
x = (2₂) bc
(A-XI)x=0
X 2 = S
to
There fore
Vector is
A
Corresponding to A X1 = -2
eigen
= (A -51) X = 0 [°° >₂=5]
(625 $35) (2) (8)
=1
-1 (1-2-3) (62) -(8
=1 x₁ +4x₂ = 0 and -22₁ -8 2₂=0
=1
IC₁ + 4x₂ = 0 and x₁ 4x2 = 0
=1
хсi + Чх 2 : 0
On solving, we get 20₁=4 and 2² = 1
Thus we
Can
take
V₂ = (-1) as
an
Vector of A Corresponding to
x = 6x + 4y
y = -2x - 3y
on
X₁ = -2 is an eigen Value of A
V2 >
answer above.
eigen Vector
25 Then
of Corresponding
(e)
Use your answer to part (a) to find the general solution of the following
system of linear differential equations, writing the solution as an
equation for x and an equation for y:
and Corresponding eigen
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