10:37 01 -1 02-2 R3 → R3 -2R₂ 1 0 -2 0 1 -1 00 V₁ X2 YO x3 thus, we get x1 - 2x3 = 0 - x₂x3 = 0 let x₁ = 2t x₂ = t = 2t 4 90-0 0 2 -1. 1 t
10:37 01 -1 02-2 R3 → R3 -2R₂ 1 0 -2 0 1 -1 00 V₁ X2 YO x3 thus, we get x1 - 2x3 = 0 - x₂x3 = 0 let x₁ = 2t x₂ = t = 2t 4 90-0 0 2 -1. 1 t
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
How did we get from the matrix above to the matrix in the bottom in part c?
Transcribed Image Text:10:37
0 1 -1
02 -2
R3 → R3 -2R₂
10
-2
6:30-0
V₁ =
1
thus, we get
x₁ - 2x3 = 0
x₂x3 = 0
let
x₁ = 2t
x₂ = t
Step3
c)
for
1
X2
:) (²)-(8)
X3
1
R₁ →
2t
t
√бi R₁
6
-1
бi
2
√бi
1
30-0
J) ()-(
-2
=
√6i
6
X1
1/²¹) (0)
√x
Do
8
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