How do you solve the system $3 x + 2 y - 3 z = - 2$ $3x+2y-3z=-2$ , $7 x - 2 y + 5 z = - 14$ $7x-2y+5z=-14$ , $2 x + 4 y + z = 6$ $2x+4y+z=6$ ?

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Answer 1 · 2018-03-13T21:02:29

$x = - \frac{22}{13}$ $x=-22/13$ , $y = \frac{29}{13}$ $y=29/13$ and $z = \frac{6}{13}$ $z=6/13$

Perform the Gauss Jordan elimination on the augmented matrix

$A=((3,2,-3,|,-2),(7,-2,5,|,-14),(2,4,1,|,6))$

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

$R1larrR1-R3$ ; $R2larrR2-3R3$

$A=((1,-2,-4,|,-8),(1,-14,2,|,-32),(2,4,1,|,6))$

$R2larrR2-R1$ ; $R3larrR3-2R1$

$A=((1,-2,-4,|,-8),(0,-12,6,|,-24),(0,8,9,|,22))$

$R3larrR3+2/3R2$

$A=((1,-2,-4,|,-8),(0,-12,6,|,-24),(0,0,13,|,6))$

$R3larr(R3)/13$

$A=((1,-2,-4,|,-8),(0,-12,6,|,-24),(0,0,1,|,6/13))$

$R1larrR1+4R3$ ; $R2larrR2-6R3$

$A=((1,-2,0,|,-80/13),(0,-12,0,|,-348/13),(0,0,1,|,6/13))$

$R2larr(R2)/(-12)$

$A=((1,-2,0,|,-80/13),(0,1,0,|,29/13),(0,0,1,|,6/13))$

$R1larrR1+2R2$

$A=((1,0,0,|,-22/13),(0,1,0,|,29/13),(0,0,1,|,6/13))$

Thus $x=-22/13$ , $y=29/13$ and $z=6/13$

How do you solve the system 3x+2y-3z=-23x+2y−3z=−23x+2y-3z=-2, 7x-2y+5z=-147x−2y+5z=−147x-2y+5z=-14, 2x+4y+z=62x+4y+z=62x+4y+z=6?