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x +y +3z=1 4x +4y +12z=4 4x +4y +12z=4 Find x, y, and z.
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but we get 0=0 for both
this means that y and z are free variables
that means that you x +y +3z=1 you can pick any values for z and y then solve for x
because it has at least one free variable it has an infinite number of solutions (this comes from the fact that you can pick any values for z and y)
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sorry I mean row 2 and 3 are scalar multiple of row 1
yea i see on this problem thank you very much these are giving me a really hard time
yup after doing a couple of them you get a hang of it...jsut keep doing problems.
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