Question

How do you solve `x+3z=-2`, `2x+2y+z=4`, and `3x+y-2z=5` using matrices?

Answer 1

`color(blue)(x=8/17" : "y=33/17" ; "z=-14/17)`

Answer approach sheet 1 of 2

Explanation:

These are a blighter to format!!!!

`" "xcolor(white)(.)ycolor(white)(.)zcolor(white)(.)" | "ans"`
`" "[(1,0,3" | "-2),(2,2,1" | "4),(3,1,-2" | "5)]`
`" "Row2-2(Row1)`
`" "Row3-3(Row1)`
`" "darr`

`color(white)(.)`
`" "[(1,0,3" | "-2),(0,2,-5" | "8),(0,1,-11" | "11)]`
`" "Row2-2(Row3)`
`" "darr`

`color(white)(.)`
`" "[(1,0,3" | "-2),(0,0,17" | "-14),(0,1,-11" | "11)]`
`" "Row2-:17`
`" "darr`

`color(white)(.)`
`" "[(1,0,3" | "-2),(0,0,1" | "-14/17),(0,1,-11" | "11)]`
`" "Row1 - 3(Row2)`
`" "darr`

`color(white)(.)`
`" "[(1,0,0" | "8/17),(0,0,1" | "-14/17),(0,1,-11" | "11)]`
`" "Row3+11(Row2)`
`" "darr`

`color(white)(.)`
`" "[(1,0,0" | "8/17),(0,0,1" | "-14/17),(0,1,0" | "33/17)]`
`" Swap "Row3" & "Row2`
`" "darr`

`color(white)(.)`
`" "[(1,0,0" | "8/17),(0,1,0" | "33/17),(0,0,1" | "-14/17)]`

`color(blue)(x=8/17" : "y=33/17" ; "z=-14/17)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check:

Answer 2

Answer approach sheet 2 of 2
Using Linear Algebra
This is a guide only to get you started

Explanation:

Let `A =([1,0,3],[2,2,1],[3,1,-2] )`

Let `B=([-2],[4],[5])`

Then `A^(-).B=([x],[y],[z])`
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
You will have to investigate how to obtain `A^(-)` as I can not remember but I can give you the matrix. There is an algorithm. The hard way is to use Gauss Jordon elimination on the LHS of the matrix below and the inverse will be in the RHS

`([1,0,3,"|",1,0,0],[2,2,1,"|",0,1,0],[3,1,2,"|",0,0,1] )`