#What is wrong with my solution method?

14 messages · Page 1 of 1 (latest)

rose condorBOT
quick mirage
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what did you do to arrive at that?

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did you row reduce the matrix?

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IMO the easiest way to approach this problem is to realize that it's really only 1 equation 3 unknowns since all 3 equations are just multiples of one another.

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So you have 5x_1 -2x_2 +2x_3 = 0

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then just divide by 5 so you have a coefficient of 1 for your first variable and solve algebraically

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letting x_2 and x_3 be free variables

sturdy lynx
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I first rescaled row 2 and row 3. When I did that I got 5, -2, and 6 for the coefficients for the 3 variables for every row.

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From there I multiplied row 1 by -1 and added the results to row 2 and 3.

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This left me with only row one being 5, -2, 6 | 0

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Row 2 and 3 resulted in 0 = 0

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And also from there, you say x2 and x3 are free variables.

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From there I just solved for x1 in terms of x2 and x3 in the first row... howver the solution that I got was off by a little.