Question

How do you solve `-3x + 5y + z = 10`, `2x + 3y - z = 7`, and `-4x + 2y +3z = -1` using matrices?

Answer 1

x=-1, y=2, z= -3

Explanation:

In matrix form, the system can be written as follows and the solved using indicated row operations:

`[(-3,5,1),(2,3,-1),(-4,2,3)][(x),(y),(z)] = [(10),(7),(-1)]`

`R_1 + R_2 to R_1`

`[(-1,8,0),(2,3,-1),(-4,2,3)][(x),(y),(z)] = [(17),(7),(-1)]`

`R_3 + 2R_3 to R_3`

`[(-1,8,0),(2,3,-1),(0,8,1)][(x),(y),(z)] = [(17),(7),(13)]`

`2R_1 + R_2 to R_2`

`[(-1,8,0),(0,19,-1),(0,8,1)][(x),(y),(z)] = [(17),(41),(13)]`

`R_2 - 2R_3 to R_2`

`[(-1,8,0),(0,3,-3),(0,8,1)][(x),(y),(z)] = [(17),(15),(13)]`

`R_2/3 to R_2`

`[(-1,8,0),(0,1,-1),(0,8,1)][(x),(y),(z)] = [(17),(5),(13)]`

`R_3 - 8R_2 to R_3`

`[(-1,8,0),(0,1,-1),(0,0,9)][(x),(y),(z)] = [(17),(5),(-27)]`

`R_3/9 to R_3`

`[(-1,8,0),(0,1,-1),(0,0,1)][(x),(y),(z)] = [(17),(5),(-3)]`

`-1R_1 to R_1`

`[(1,-8,0),(0,1,-1),(0,0,1)][(x),(y),(z)] = [(-17),(5),(-3)]`

`R_3 + R_2 to R_2`

`[(1,-8,0),(0,1,0),(0,0,1)][(x),(y),(z)] = [(-17),(2),(-3)]`

`8R_2 + R_1 to R_1`

`[(1,0,0),(0,1,0),(0,0,1)][(x),(y),(z)] = [(-1),(2),(-3)]`