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

1 Answer
Dec 29, 2016

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)]

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