Question

How do you solve using gaussian elimination or gauss-jordan elimination, `5x + y + 5z = 3`, `4x − y + 5z = 13`, `5x + 2y + 2z = 2`?

Answer 1

`x=1 32/99`,
`y=-5 131/198`,
`z=9/22`

Explanation:

To solve the system of linear equations

`{(5x+y+5z=3),(4x-y+5z=13),(5x+2y+2z=2):}`

First convert it to the augmented matrix form

`Rightarrow [(5,1,5,|,3),(4,-1,5,|,13),(5,2,2,|,2)]`

The goal is to write this in row-echelon form, which is triangular form where the leading coefficient is 1. The three row operations we can use are:

1. Switch rows
2. Multiply a row by a constant
3. Add a multiple of a row to another

Multiply the first row by `1/5`

`Rightarrow [(1,1/5,1,|,3/5),(4,-1,5,|,13),(5,2,2,|,2)]`

Multiply first row by -4 and add to second row

`Rightarrow [(1,1/5,1,|,3/5),(0,-9/5,1,|,53/5),(5,2,2,|,2)]`

Multiply first row by -5 and add to the third row

`Rightarrow [(1,1/5,1,|,3/5),(0,-9/5,1,|,53/5),(0,1,-3,|,-1)]`

Multiply row 2 by `-5//9`

`Rightarrow [(1,1/5,1,|,3/5),(0,1,-5/9,|,-53/9),(0,1,-3,|,-1)]`

Multiply row 2 by -1 and add to row 3

`Rightarrow [(1,1/5,1,|,3/5),(0,1,-5/9,|,-53/9),(0,0,-22/9,|,-1)]`

Multiply the last row by -9/22

`Rightarrow [(1,1/5,1,|,3/5),(0,1,-5/9,|,-53/9),(0,0,1,|,9/22)]`

Plugging these values back into formulas gives

`{(x+1/5y+z=3/5),(y-5/9z=-53/9),(z=9/22):}`

Using backwards substitution we get the remaining answers:
`x=131/99=1 32/99`,
`y=-1121/198=-5 131/198`,
`z=9/22`