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I've just started reading a couple of books about Linear Algebra and there is a simple explanation on one of them that confused me. The author analyses a system of two equations with two variables. He calculates the solution asuming that all the constants "a" and "b" are different from 0. Texbook Example This I understand. If the denominator in the last step is 0, the unique x that satisfies both equations does not exist, because the slope of both equations is the same. There's either no solution of infinite solutions. My problem is that when messing around I can also end up with this: My Example This is clearly wrong. I checked just to make sure, and it was, but I can't figure out why. What's the difference between the two calculations? I mean, both are using the same x/y values, right? I know I would never use the latter formula because it still has two variables defeating the purpose of solving the system. What confuses me is what does it mean that when "a11-a21" equals to 0 the system's solution is undetermined, when it may actually be a solution. Did I screw up somewhere? I got that the conclusion it is only true when the constants multiplied by "y" on both equations are the same, but I'm not sure what to make of it.

Thank you for your time.

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The difference is that in the first example $x$ is given in terms of just $a$'s and $b$'s. These are actual numbers, not variables you are trying to solve for. In your example, you have an equality between $x$ and a not-yet-shown-to-exist $y$.

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  • $\begingroup$ But if "a11-a21" were 0, wouldn't the "y" value be irrelevant? Is that the problem? That I'm disregarding/overriding the "y" when assuming what values the constants should have for the system to have no unique solution? $\endgroup$
    – user219634
    Feb 27, 2015 at 4:15
  • $\begingroup$ If $a_{11}-a_{21}$ were zero, you couldn't have divided by it. You would just be left with $(a_{12}-a_{22})y = ...$ which tells you what $y$ must be as long as $a_{12}-a_{22}$. Even if you know what $y$ must be though, your choice of $x$ is free. $\endgroup$
    – nullUser
    Feb 27, 2015 at 4:23

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