Question

How do you solve the below equation using reduced row-echelon form?

`2x+3y-4z=9`
`x+y+z=3`
`2x+y-3z=2`

Answer 1

`x = -7/11,y = 39/11, z = 1/11`

Explanation:

Start with the augmented matrix `(A|b)`

`((2,3,-4,|,9),(1,1,1,|,3),(2,1,-3,|,2))`

We carry out row reductions to transform this to the reduced row-echelon form. We first swap rows 1 and 2 (this is not necessary - but will cut down some of the computation)

`((1,1,1,|,3),(2,3,-4,|,9),(2,1,-3,|,2))`

`R_2 -> R_2-2R_1, R_3->R_3-2R_1`
`((1,1,1,|,3),(0,1,-6,|,3),(0,-1,-5,|,-4))`

`R_1->R_1-R_2, R_3->R_3+R_2`
`((1,0,7,|,0),(0,1,-6,|,3),(0,0,-11,|,-1))`

`R_3 -> R_3/{-11}`
`((1,0,7,|,0),(0,1,-6,|,3),(0,0,1,|,1/11))`

`R_1 -> R_1-7R_3, R_2 -> R_2+6R_3`
`((1,0,0,|,-7/11),(0,1,0,|,39/11),(0,0,1,|,1/11))`

This gives
`x = -7/11,y = 39/11, z = 1/11`