How do you solve the system $x - 4 y + 4 z = - 1$ $x-4y+4z=-1$ , $y - 3 z = 5$ $y-3z=5$ , and $3 x - 4 y + 6 z = 1$ $3x-4y+6z=1$ ?

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Answer 1 · 2018-03-12T01:33:34

$x = 3$ $x=3$ , $y = - 1$ $y=-1$ and $z = - 2$ $z=-2$

Perform the Gauss Jordan elimination on the augmented matrix

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

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

$R3larrR3-3R1$

$A=((1,-4,4,|,-1),(0,1,-3,|,5),(0,8,-6,|,4))$

$R1larrR1+4R2$ ; $R3larrR3-8R2$

$A=((1,0,-8,|,19),(0,1,-3,|,5),(0,0,18,|,-36))$

$R3larr(R3)/18$

$A=((1,0,-8,|,19),(0,1,-3,|,5),(0,0,1,|,-2))$

$R1larrR1+8R3$ ; $R2larrR2+3R3$

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

Thus $x=3$ , $y=-1$ and $z=-2$

How do you solve the system x-4y+4z=-1x−4y+4z=−1x-4y+4z=-1, y-3z=5y−3z=5y-3z=5, and 3x-4y+6z=13x−4y+6z=13x-4y+6z=1?