Question

Solving a linear system? x+2y+z=2 3x+8y+z=12 4y+z=2

Answer 1

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

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

`A=((1,2,1,|,2),(3,8,1,|,12),(0,4,1,|,2))`

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`

`A=((1,2,1,|,2),(0,2,-2,|,6),(0,4,1,|,2))`

`R3larrR3-2R2`

`A=((1,2,1,|,2),(0,2,-2,|,6),(0,0,5,|,-10))`

`R3larr(R3)/5`

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

`R1larrR1-R3`; `R2larrR2+2R3`

`A=((1,2,0,|,4),(0,2,0,|,2),(0,0,1,|,-2))`

`R1larrR1-R2`;

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

`R2larr(R2)/2`

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