The system of linear equations is,

`x-y-4z = 12`

`2x+y+3z = -6`

`5x+2y+2z = 1`

The augmented matrix is, (I will write rows separated from a comma, as I am unable to type matrix here)

(1 -1 -4, 2 1 3, 5 2 2) = (12,-6,1)

And I will perform...

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The system of linear equations is,

`x-y-4z = 12`

`2x+y+3z = -6`

`5x+2y+2z = 1`

The augmented matrix is,

(I will write rows separated from a comma, as I am unable to type matrix here)

(1 -1 -4, 2 1 3, 5 2 2) = (12,-6,1)

And I will perform matrix operations to produce the row echelon form.

R2 = -R1 x 2 + R2

(1 -1 -4, 0 3 11, 5 2 2) = (12,-30, 1)

R3 = -R1 x 5 + R3

(1 -1 -4, 0 3 11, 0 7 22) = (12,-30, -59)

R2 = R2 / 3 and R3 = R3 / 7

(1 -1 -4, 0 1 11/3, 0 1 22/7) = (12,-10, -59/7)

R3 = -R2+R3

(1 -1 -4, 0 1 11/3, 0 0 -11/21) = (12,-10, 11/7)

R3 = -(21/11) R3

(1 -1 -4, 0 1 11/3, 0 0 1) = (12,-10, -3)

This is the reduced row echelon form.

This gives us,

`z = -3`

and

`y+11/3z = -10`

`y+11/3(-3) =-10`

`y -11=-10`

`y = 1`

and finally from first row,

`x-y-4z = 12`

`x -1+12 = 12`

`x =1`

**The answers are x= 1, y = 1 and z =-3 or [1,1,-3]**