Using Gauss-Jordan elim solve x + y + x =0, 2x -y +z=1,and x + y -2z = 2?

Question

explain

Answer 1

The solution in matrix form:

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The matrices represent these steps:

Initial system:

x+y+z = 0

2x-y+z = 1

x+y-2z = 2

Eliminate x terms from 2nd & 3rd equations:

x+y+z = 0

-3y-z = 1

-3z = 2

x+y+z = 0

y+(1/3)z = -1/3   coefficient of y term set to 1

-3z = 2

Eliminate y terms from 1st & 3rd equations:

x+(2/3)z = 1/3

y+(1/3)z = -1/3

-3z = 2

x+(2/3)z = 1/3

y+(1/3)z = -1/3

z = -2/3   coefficient of z term set to 1

Eliminate z terms from 1st & 2nd equations:

x = 7/9

y = -1/9

z = -2/3

Answer 2

wait. Is the first equation suppose to be "x+y+x = 0" or "x+y+z =0"

Please advise.  

Answer 3

setup matrix and convert to reduced-row echelon form

which is in form of

1 0 0 x

0 1 0 y

0 0 1 z

so our matrix is

1 1 1 0

2 -1 1 1

1 1 -2 2

multiply first row by -2 then add it to second row and apply it to second row

-2R(1) + R(2) → R(2)

1 1 1 0

0 -3 -1 1

1 1 -2 -2

now multiply first row by -1 and add it to third row and apply it to third row

-1R(1) + R(3) → R(3)

1 1 1 0

0 -3 -1 1

0 0 -3 2

now divide row 3 by -3

R(3) / -3 → R(3)

1 1 1 0

0 -3 -1 1

0 0 1 -2/3

now add row 3 to row 2 and apply it to row 2

R(3) + R(2) → R(2)

1 1 1 0

0 -3 0 1/3

0 0 1 -2/3

now divide row 2 by -3

R(2) / -3 → R(2)

1 1 1 0

0 1 0 -1/9

0 0 1 -2/3

now multiply row 3 by -1 and add it to row 1 and apply it to row 1

-1R(3) + R(1) → R(1)

1 1 0 2/3

0 1 0 -1/9

0 0 1 -2/3

now final step is to multiply row 2 by -1 and add it to row 1 and apply it to row 1

1 0 0 7/9

0 1 0 -1/9

0 0 1 -2/3

and now the matrix is in reduced row echelon form and we can simply read off the answers

x= 7/9    y= -1/9    z = -2/3

hope this helps