Question

How do you solve x+y−z=6, 2x−y+z=−9, and x−2y+3z=1 using matrices?

Answer 1

x+y-z=6

2x-y+z=-9

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3x=3

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x=1

so1+y-z=6

y-z=5----------1

2-y+z=-9

z-y=-7----------2

x-2y+3z=1

1-2y+3z=1

3z-2y=0---------3

from eq2

z=-7+y

substitute in eq3

3(-7+y)-2y=03z

-21+3y-2y=0

y=21

substitute y value in eq3

3z-42=0

3z=42

z=14