How do you solve the system $x+3y-2z=8$ , $3x+2y-3z=15$ , and $4x+2y+3z=-1$ ?

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Answer 1 · 2018-01-03T19:32:54

$x=134/47$ , $y=24/47$ and $z=-85/47$

Perform the Gauss Jordan elimination on the augmented matrix

$A=((1,3,-2,|,8),(3,2,-3,|,15),(4,2,3,|,-1))$

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

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

$A=((1,3,-2,|,8),(0,-7,3,|,-9),(0,-10,11,|,-25))$

$R2larrR2-R3$

$A=((1,3,-2,|,8),(0,3,-8,|,16),(0,-10,11,|,-25))$

$R1larrR1-R2$

$A=((1,0,6,|,-8),(0,3,-8,|,16),(0,-10,11,|,-25))$

$R2larr(R2)/3$

$A=((1,0,6,|,-8),(0,1,-8/3,|,16/3),(0,-10,11,|,-25))$

$R3larrR3+10R2$

$A=((1,0,6,|,-8),(0,1,-8/3,|,16/3),(0,0,-47/3,|,85/3))$

$R3larrR3*(-3)/47$

$A=((1,0,6,|,-8),(0,1,-8/3,|,16/3),(0,0,1,|,-85/47))$

$R1larrR1-6R3$ ; $R2larrR2+8/3R3$

$A=((1,0,0,|,134/47),(0,1,0,|,24/47),(0,0,1,|,-85/47))$

Thus $x=134/47$ , $y=24/47$ and $z=-85/47$

How do you solve the system x+3y-2z=8x+3y-2z=8, 3x+2y-3z=153x+2y-3z=15, and 4x+2y+3z=-14x+2y+3z=-1?