How do you solve $x + y + z = 3$ , $-x + 3y + 2z = -8$ , and $5y + z = 2$ ?

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Answer 1 · 2018-01-13T22:38:37

$x=5$ , $y=1$ and $z=-3$

Perform the Gauss Jordan elimination on the augmented matrix

$A=((1,1,1,|,3),(-1,3,2,|,-8),(0,5,1,|,2))$

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

$R2larrR2+R1$ ;

$A=((1,1,1,|,3),(0,4,3,|,-5),(0,5,1,|,2))$

$R2larrR2*5$ ; $R3larrR3*4$

$A=((1,1,1,|,3),(0,20,15,|,-25),(0,20,4,|,8))$

$R3larrR3-R2$

$A=((1,1,1,|,3),(0,20,15,|,-25),(0,0,-11,|,33))$

$R3larr(R3)/(-11)$

$A=((1,1,1,|,3),(0,20,15,|,-25),(0,0,1,|,-3))$

$R1larrR1-R3$ ; $R2larrR2-15R3$

$A=((1,1,0,|,6),(0,20,0,|,20),(0,0,1,|,-3))$

$R2larr(R2)/20$

$A=((1,1,0,|,6),(0,1,0,|,1),(0,0,1,|,-3))$

$R1larrR1-R2$

$A=((1,0,0,|,5),(0,1,0,|,1),(0,0,1,|,-3))$

Thus $x=5$ , $y=1$ and $z=-3$

How do you solve x + y + z = 3x + y + z = 3, -x + 3y + 2z = -8-x + 3y + 2z = -8, and 5y + z = 25y + z = 2?