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How would you solve the following system of linear equations: $2x_1 + (3+a)x_2 + 2x_3 = 2+a$

$x_1 + ax_2 + 2x_3 = a$

$ax_1 + 2x_2 + 2ax_3 = 0$

assuming that $a \neq \pm \sqrt{2}$? I feel confident in solving linear equation systems with just constants as the coefficients but the variable coefficient $a$ is what gives me problems.

Here is the solution I find: solution as augmented matrix but how would the assumption about a make the reduced echelon form any different compared to an assumption about e.g. $a = \pm \sqrt{2}$?

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  • $\begingroup$ Have you tried Gaussian elimination? $\endgroup$ Commented Apr 28, 2019 at 16:31
  • $\begingroup$ @saulspatz Yes and I do get a solution set that includes a but why is the "assume that a is..." important for this equation system? I mean, why is it necessary to assume something about a if there is a general solution set anyway? $\endgroup$ Commented Apr 28, 2019 at 16:54
  • $\begingroup$ Did you ever divide by $a^2-2$? That's the most obvious reason for the condition. $\endgroup$ Commented Apr 28, 2019 at 16:56
  • $\begingroup$ @saulspatz Thanks for taking the time to help me :) I updated the OP a little with my found solution set. I never used the assumption about a for any EROs because but shouldn't it also work if you applied it AFTER you found the general solution set? Please see my updated question (last part) $\endgroup$ Commented Apr 28, 2019 at 17:05
  • $\begingroup$ No. This system is inconsistent if $a=\pm\sqrt2$. More generally, the rank of the coefficient matrix can be different for different values of any variable coefficients, so you have to analyze the system separately for each exceptional case. $\endgroup$ Commented Apr 28, 2019 at 17:24

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With $$x_1=a-ax_2-2x_3$$ can we eliminate $$x_1$$: $$x_2(3-a)-2x_3=2-a$$ $$x_2(2-a^2)+4ax_3=-a^2$$ And then using $$x_3=\frac{x_2(3-a)-(2-a)}{2}$$ to eliminate $$x_3$$. Then you will get $$x_2(2-a^2)+2a(3-a)x_2-2a(2-a)=-a^2$$

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  • $\begingroup$ Thanks but how are you using the assumption that $a \neq \pm \sqrt{2}$? I have been told that the solution set for the equation system is different when assuming that $a \neq \pm \sqrt{2}$ and $a = \pm \sqrt{2}$. $\endgroup$ Commented Apr 28, 2019 at 17:20
  • $\begingroup$ I was not ready yet. $\endgroup$ Commented Apr 28, 2019 at 17:25
  • $\begingroup$ Ahh sry haha :) I'll get some coffee and come back in a little then :) $\endgroup$ Commented Apr 28, 2019 at 17:26
  • $\begingroup$ Where did you get the solutions for all the x variables? $\endgroup$ Commented Apr 28, 2019 at 18:18
  • $\begingroup$ From your system, at first $$x_1$$, then $$x_3$$ and we get an equation only in $$x_2$$ including the parameters. $\endgroup$ Commented Apr 28, 2019 at 18:21

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