Solve using Gaussian Elimination with back-substitution or Gauss Jordan 3z1 - 2x2 +4x3 1 21+2-2x3 = 3 %3D 221 – 3r2 +6x3 8 O 21 = 2, 22 = 4, 23 = 6 %3D %3D %3D O 21 = -3, 2 = -5, a3 = -7 %3D %3D O z1 = 3, 2 = 5, 23 7 %3D %3D %3D No solution

Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter11: Matrices And Determinants
Section11.CT: Chapter Test
Problem 6CT
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Solve using Gaussian Elimination with back-substitution or Gauss Jordan
3z1 – 2x2 +4x3 = 1
2i + x2 - 2x3 = 3
%3D
221 – 3z2 + 6x3 = 8
%3D
O 21 = 2, 22 = 4, 13 = 6
%3D
%3D
O 21 = -3, 2 = -5, 23 = -7
%3D
%3D
21 = 3, 22 = 5, 23 = 7
%3D
%3D
%3D
No solution
Transcribed Image Text:Solve using Gaussian Elimination with back-substitution or Gauss Jordan 3z1 – 2x2 +4x3 = 1 2i + x2 - 2x3 = 3 %3D 221 – 3z2 + 6x3 = 8 %3D O 21 = 2, 22 = 4, 13 = 6 %3D %3D O 21 = -3, 2 = -5, 23 = -7 %3D %3D 21 = 3, 22 = 5, 23 = 7 %3D %3D %3D No solution
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