Question

How do I solve this system of linear equations? `x+y+z=-1`, `2x+4y+z=11`, `x+2y-3z=-3`

Does anyone know how to solve a system of equations? if so can you please help me with this one?
`x+y+z=-1`
`2x+4y+z=11`
`x+2y-3z=-3`

Answer 1

`x=-78/7`, `y=54/7` and `z=17/7`

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

`A=((1,1,1,|,-1),(2,4,1,|,11),(1,2,-3,|,-3))`

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

`R2larrR2-2R1`; `R3larrR3-R1`

`A=((1,1,1,|,-1),(0,2,-1,|,13),(0,1,-4,|,-2))`

`R3larrR3*2`

`A=((1,1,1,|,-1),(0,2,-1,|,13),(0,2,-8,|,-4))`

`R3larrR3-R2`

`A=((1,1,1,|,-1),(0,2,-1,|,13),(0,0,-7,|,-17))`

`R3larr(R3)/(-7)`

`A=((1,1,1,|,-1),(0,2,-1,|,13),(0,0,1,|,17/7))`

`R1larrR1-R3`; `R2larrR2+R3`

`A=((1,1,0,|,-24/7),(0,2,0,|,108/7),(0,0,1,|,17/7))`

`R2larr(R2)/2`

`A=((1,1,0,|,-24/7),(0,1,0,|,54/7),(0,0,1,|,17/7))`

`R1larrR1-R2`

`A=((1,0,0,|,-78/7),(0,1,0,|,54/7),(0,0,1,|,17/7))`

Thus `x=-78/7`, `y=54/7` and `z=17/7`