Solve the system of equation below. Modulus of each unknown. Given : 3u + (1 – i)v+ (3 + 4i)w = (46 + v3)(1 – i) +i(66 – 2/3 %3D (5+ 2i)u – (6 + 2i)v – (2 + 5i)w = (-34 – 6v3) + i(66 – 2v3) 6V3 3iu + 5v + 2(1 + i)w= 5(1+ V3) – 17i Find: |w| – |v| – |u|.
Solve the system of equation below. Modulus of each unknown. Given : 3u + (1 – i)v+ (3 + 4i)w = (46 + v3)(1 – i) +i(66 – 2/3 %3D (5+ 2i)u – (6 + 2i)v – (2 + 5i)w = (-34 – 6v3) + i(66 – 2v3) 6V3 3iu + 5v + 2(1 + i)w= 5(1+ V3) – 17i Find: |w| – |v| – |u|.
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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Question
![Solve the system of equation
below. Modulus of each unknown.
Given :
Зи + (1 — і)о + (3+ 4i)w 3 (4
+ v3) (1 – i)
(5+ 21)и — (6 + 2:)u — (2 + 5і)w %3 (-3
6V3
– 2v3)
+i(66 –
|
3iu + 5v + 2(1 + i)w= 5(1+ V3)
17i
|
Find: |w| – |v| – |u].
Your answer](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F86c3cc99-4742-436c-a507-c734d16d568b%2F75506ac5-c34e-40e8-b15a-3dce67290cb6%2Fyuh06od_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Solve the system of equation
below. Modulus of each unknown.
Given :
Зи + (1 — і)о + (3+ 4i)w 3 (4
+ v3) (1 – i)
(5+ 21)и — (6 + 2:)u — (2 + 5і)w %3 (-3
6V3
– 2v3)
+i(66 –
|
3iu + 5v + 2(1 + i)w= 5(1+ V3)
17i
|
Find: |w| – |v| – |u].
Your answer
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