Solve by Gauss-Jordan elimination

Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter9: Systems Of Linear Equations
Section9.2: The Substitution Method
Problem 32WE
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Solve the problem with a complete and detailed solution.

 

 

Solve by Gauss-Jordan elimination.
х + 3x, — 2хз
+ 2x5
2х1 + 6х, — 5х; — 2х, + 4x;
3x, = -1
-
+ 15x, =
+ 8x4 + 4x5 + 18x, =
5x3 + 10x4
2x1 + 6x2
Problem 2:
Show that the following nonlinear system has 18 solutions if
0< a < 27,0 < ß < 27, and 0 < y < 277.
sin a + 2 cos B + 3 tan y = 0
2 sin a + 5 cos ß + 3 tan y = 0
- sin a – 5 cos 3 + 5 tan y = 0
[Hint: Begin by making the substitutions x =
y = cos ß, and z = tan y.]
sin a,
Transcribed Image Text:Solve by Gauss-Jordan elimination. х + 3x, — 2хз + 2x5 2х1 + 6х, — 5х; — 2х, + 4x; 3x, = -1 - + 15x, = + 8x4 + 4x5 + 18x, = 5x3 + 10x4 2x1 + 6x2 Problem 2: Show that the following nonlinear system has 18 solutions if 0< a < 27,0 < ß < 27, and 0 < y < 277. sin a + 2 cos B + 3 tan y = 0 2 sin a + 5 cos ß + 3 tan y = 0 - sin a – 5 cos 3 + 5 tan y = 0 [Hint: Begin by making the substitutions x = y = cos ß, and z = tan y.] sin a,
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