How do you solve the system $x+2y-4z=13$ , $3x+4y-2z=19$ , and $3x+2z=3$ ?

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Answer 1 · 2017-12-16T13:00:28

$x=2$ , $y=5/2$ and $z=-3/2$

Perform the Gauss Jordan elimination on the augmented matrix

$A=((1,2,-4,|,13),(3,4,-2,|,19),(3,0,2,|,3))$

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

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

$A=((1,2,-4,|,13),(0,-2,10,|,-20),(0,-6,14,|,-36))$

$R3larrR3-3R2$

$A=((1,2,-4,|,13),(0,-2,10,|,-20),(0,0,-16,|,24))$

$R3larr(R3)/(-16)$

$A=((1,2,-4,|,13),(0,-2,10,|,-20),(0,0,1,|,-3/2))$

$R1larrR1+4R3$ ; $R2larrR2-10R3$

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

$R1larrR1+R2$

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

$R2larr(R2)/(-2)$

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

Thus, solution of equation system is $x=2$ , $y=5/2$ and $z=-3/2$

How do you solve the system x+2y-4z=13x+2y-4z=13, 3x+4y-2z=193x+4y-2z=19, and 3x+2z=33x+2z=3?