How do you solve the system $x-y-2z=-6$ , $3x+2y=-25$ , and $-4x+y-z=12$ ?

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Answer 1 · 2018-02-28T19:17:51

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

Perform the Gauss Jordan elimination on the augmented matrix

$A=((1,-1,-2,|,-6),(3,2,0,|,-25),(-4,1,-1,|,12))$

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

$R2larrR2-3R1$ , $R3larrR3+4R1$ ;

$A=((1,-1,-2,|,-6),(0,5,6,|,-7),(0,-3,-9,|,-12))$

$R3larr(R3)/(-3)$ ;

$A=((1,-1,-2,|,-6),(0,5,6,|,-7),(0,1,3,|,4))$

$R1larrR1+R3$ , $R2larrR2-5R3$ ;

$A=((1,0,1,|,-2),(0,0,-9,|,-27),(0,1,3,|,4))$

$R2larr(R2)/(-9)$ ;

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

$R1larrR1-R2$ , $R3larrR3-3R2$ ;

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

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

How do you solve the system x-y-2z=-6x-y-2z=-6, 3x+2y=-253x+2y=-25, and -4x+y-z=12-4x+y-z=12?