Question

How do you solve using gaussian elimination or gauss-jordan elimination, `y + 3z = 6`, `x + 2y + 4z = 9`, `2x + y + 6z = 11`?

Answer 1

Using Gaussian elimination
`color(white)("XXX")(x,y,z)= (1/7,9/7,11/7)`

Explanation:

Converting the given equations to augmented matrix (to avoid having to re-write the variable names)
`{(0,1,3,6),(1,2,4,9),(2,1,6,11):}color(white)("XX"){:([1]),([2]),([3]):}`

Exchanging rows [1] and [2]
`{(1,2,4,9),(0,1,3,6),(2,1,6,11):}color(white)("XX"){:([4]),([5]),([3]):}`

Note that at this point rows [4] and [5] are in Gaussian form
and all that remains is to zero out the first two columns of row [3]

Subtracting `2` times row [4] from row [3]
`{(1,2,4,9),(0,1,3,6),(0,-3,-2,-7):}color(white)("XX"){:([4]),([5]),([6]):}`

Adding `3` times row [5] to row [6]
`{(1,2,4,9),(0,1,3,6),(0,0,7,11):}color(white)("XX"){:([4]),([5]),([7]):}`

Dividing row [7] by `7`
`{(1,2,4,9),(0,1,3,6),(0,0,1,11/7):}color(white)("XX"){:([4]),([5]),([8]):}`

This gives us
`color(white)("XXX")z=11/7`

Substituting back into row [5]
`color(white)("XXX")y+3xx11/7=6 rarr y= 9/7`

Substituting back into row [4]
`color(white)("XXX")x+2xx9/7+4xx11/7=9 rarr x=1/7`