Solving a linear system? x+2y+z=2 3x+8y+z=12 4y+z=2

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Answer 1 · 2018-07-04T10:49:08

$x = 2$ $x=2$ , $y = 1$ $y=1$ and $z = - 2$ $z=-2$

Perform the Gauss Jordan elimination on the augmented matrix

$A=((1,2,1,|,2),(3,8,1,|,12),(0,4,1,|,2))$

I have written the equations not in the sequence as in the question in order to get $1$ as pivot.

Perform the folowing operations on the rows of the matrix

$R2larrR2-3R1$

$A=((1,2,1,|,2),(0,2,-2,|,6),(0,4,1,|,2))$

$R3larrR3-2R2$

$A=((1,2,1,|,2),(0,2,-2,|,6),(0,0,5,|,-10))$

$R3larr(R3)/5$

$A=((1,2,1,|,2),(0,2,-2,|,6),(0,0,1,|,-2))$

$R1larrR1-R3$ ; $R2larrR2+2R3$

$A=((1,2,0,|,4),(0,2,0,|,2),(0,0,1,|,-2))$

$R1larrR1-R2$ ;

$A=((1,0,0,|,2),(0,1,0,|,1),(0,0,1,|,-2))$

$R2larr(R2)/2$

Thus $x=2$ , $y=1$ and $z=-2$