Question

How do you solve the system `-6x-2y-z=-17`, `5x+y-6z=19`, and `-4x-6y-6z=-20`?

Answer 1

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

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

`A=((-6,-2,-1,|,-17),(5,1,-6,|,19),(-4,-6,-6,|,-20))`

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+R2`; `R3larrR3+R2`

`A=((-1,-1,-7,|,2),(5,1,-6,|,19),(1,-5,-12,|,-1))`

`R2larr(R2)+5R1`; `R3larrR3+R1`

`A=((-1,-1,-7,|,2),(0,-4,-41,|,29),(0,-6,-19,|,1))`

`R1larrR1*(-1)`; `R2larrR2-R3`

`A=((1,1,7,|,-2),(0,2,-22,|,28),(0,-6,-19,|,1))`

`R3larrR3+3R2`

`A=((1,1,7,|,-2),(0,2,-22,|,28),(0,0,-85,|,85))`

`R3larr(R3)/(-85)`

`A=((1,1,7,|,-2),(0,2,-22,|,28),(0,0,1,|,-1))`

`R1larrR1-7R3`; `R2larrR1+22R3`

`A=((1,1,0,|,5),(0,2,0,|,6),(0,0,1,|,-1))`

`R2larr(R2)/2`

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

`R1larrR1-R2`

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

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