How do you solve using gaussian elimination or gauss-jordan elimination, $x + y + z = 1$ $x+y+z=1$ , $x + y - 2 z = 3$ $x+y-2z=3$ , $x + 2 y + z = 2$ $x+2y+z=2$ ?

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Answer 1 · 2018-07-11T14:36:27

The solution is $((x),(y),(y))=((2/3),(1),(-2/3))$

Perform the Jordan-Gauss elimination on the augmented matrix

$A=((1,1,1,|,1),(1,1,-2,|,3),(1,2,1,|,2))$

The pivot is in the first colum and the first row.

Eliminate the first column by performing the row operations

$R2larrR2-R1$ and $R3larrR3-R1$

$((1,1,1,|,1),(0,0,-3,|,2),(0,1,0,|,1))$

Swap the last two rows

$R2harrR3$

$((1,1,1,|,1),(0,1,0,|,1),(0,0,-3,|,2))$

The pivot is in the second row and second column

Eliminate the second column

$R1larrR1-R2$

$((1,0,1,|,0),(0,1,0,|,1),(0,0,-3,|,2))$

Make the pivot in the third row of third column

$R3larrR3*(-1/3)$

$((1,0,1,|,0),(0,1,0,|,1),(0,0,1,|,-2/3))$

Eliminate the third column

$R1larrR1-R3$

$((1,0,0,|,2/3),(0,1,0,|,1),(0,0,1,|,-2/3))$

The solution is

$((x),(y),(y))=((2/3),(1),(-2/3))$

How do you solve using gaussian elimination or gauss-jordan elimination, x+y+z=1x+y+z=1x+y+z=1, x+y-2z=3x+y−2z=3x+y-2z=3, x+2y+z=2x+2y+z=2x+2y+z=2?