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Spot the mistakes/error in solving the linear system, and then say something about it.    Given : x +  y + z = 6                   2y+5z=-4            2x+ 5y -z =4

Spot the mistakes/error in solving the linear system, and then say something about it.    Given : x +  y + z = 6                   2y+5z=-4            2x+ 5y -z =4

Elementary Linear Algebra (MindTap Course List)
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter3: Determinants
Section3.4: Applications Of Determinants
Problem 27E: Use Cramers Rule to solve the system of linear equations for x and y. kx+(1k)y=1(1k)x+ky=3 For what...
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Spot the mistakes/error in solving the linear system, and then say something about it. 

 

Given : x +  y + z = 6

                  2y+5z=-4

           2x+ 5y -z =4

Eliminate  x from 2nd  & 3rd equation :

         2x+5y-z=27

        2x+2y+2z=12

        0   3y-3z=15

And we get :

       x+y+z=6

           2y+5z= -4

          3y-3z=15

 

Eliminate y  from 3rd equation:

             6y-6z=30

            6y+15z=-12

             0-2z=42

                     z=-2

          

And we end up with:

          x + y +  z =6

                2y   + 5z =  4

                          z= -2

By back-substituting : 

     x = 5;   y= 3  and z = -2 

 

Spot the mistakes/error in solving the linear system, and then say something about it. x + y + z 6 %3D Given : 2y + 5z -4 2х + 5y 27 %3D Eliminate x from 2nd & 3rd equation : 2x + 5y - z = 27 +2z = 12 +2y O 3y -3z = 15 And we get : x + y 6 2y 5z -4 3y - 3z = 15 Eliminate y from 3rd equation: 6y - 6z = 30 6y +15z = -12 8 -2lz = 42 » z = -2 And we end up with: x+y+ z =6 2y + 5z = 4 z= -2 By back-substituting: x = 5; y= 3 and z = -2
Transcribed Image Text:Spot the mistakes/error in solving the linear system, and then say something about it. x + y + z 6 %3D Given : 2y + 5z -4 2х + 5y 27 %3D Eliminate x from 2nd & 3rd equation : 2x + 5y - z = 27 +2z = 12 +2y O 3y -3z = 15 And we get : x + y 6 2y 5z -4 3y - 3z = 15 Eliminate y from 3rd equation: 6y - 6z = 30 6y +15z = -12 8 -2lz = 42 » z = -2 And we end up with: x+y+ z =6 2y + 5z = 4 z= -2 By back-substituting: x = 5; y= 3 and z = -2
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