Question

How do you solve using gaussian elimination or gauss-jordan elimination, `2x+4y-6z=48`, `x+2y+3z=-6`, `3x-4y+4z=-23`?

Answer 1

Using Gauss elimination gives `x=3`, `y=3`, and `z=-5`

Explanation:

Step 1. Rewrite the equations in matrix form
`((2,4,-6,|,48),(1,2,3,|,-6),(3,-4,4,|,-23))`

Step 2. Switch `R_2` with `R_1`
`((1,2,3,|,-6),(2,4,-6,|,48),(3,-4,4,|,-23))`

Step 3. `-2R_1+R_2->R_2`
`((1,2,3,|,-6),(0,0,-12,|,60),(3,-4,4,|,-23))`

Step 4. `-3R_1+R_3->R_3`
`((1,2,3,|,-6),(0,0,-12,|,60),(0,-10,-5,|,-5))`

Step 5. Switch `R_2` with `R_3`
`((1,2,3,|,-6),(0,-10,-5,|,-5),(0,0,-12,|,60))`

Step 6. `-1/10R_2 -> R_2`
`((1,2,3,|,-6),(0,1,1/2,|,1/2),(0,0,-12,|,60))`

Step 7. `-1/12R_3->R_3`
`((1,2,3,|,-6),(0,1,1/2,|,1/2),(0,0,1,|,-5))`

Translating back into the values `x`, `y`, and `z`
`z=-5`
`y+1/2(-5)=1/2=>y=3`
`x+2(3)+3(-5)=-6=>x=3`