Question

How do you solve `x + 3y + z = 3`; `x + 5y + 5z = 1`; `2x + 6y + 3z = 8` using matrices?

Answer 1

The solution is `((x),(y),(z))=((16),(-5),(2))`

Explanation:

Perform the Gauss-Jordan elimination on the augmented matrix

`A=((1,3,1,|,3),(1,5,5,|,1),(2,6,3,|,8))`

Make the pivot in the first column and the first row

Eliminate the first column, perform the row operations

`R2larrR2-R1`, and `R3larrR3-2R1`

`=((1,3,1,|,3),(0,2,4,|,-2),(0,0,1,|,2))`

Make the pivot in the second column by

`R2larr(R2)/2`

`=((1,3,1,|,3),(0,1,2,|,-1),(0,0,1,|,2))`

Eliminate the second column by

`R1larrR1-3R2`

`=((1,0,-5,|,6),(0,1,2,|,-1),(0,0,1,|,2))`

Make the pivot in the second column, and eliminate the third column

`R1larrR1+5R3` and `R2larrR2-2R3`

`=((1,0,0,|,16),(0,1,0,|,-5),(0,0,1,|,2))`

The solution is

`((x),(y),(z))=((16),(-5),(2))`