# Solve linear equation system with variable real number?

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. $$a = \pm \sqrt{2}$$?

• Have you tried Gaussian elimination? Apr 28, 2019 at 16:31
• @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? Apr 28, 2019 at 16:54
• Did you ever divide by $a^2-2$? That's the most obvious reason for the condition. Apr 28, 2019 at 16:56
• @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) Apr 28, 2019 at 17:05
• 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.
– amd
Apr 28, 2019 at 17:24

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 $$x_2(2-a^2)+2a(3-a)x_2-2a(2-a)=-a^2$$
• 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 $a = \pm \sqrt{2}$. Apr 28, 2019 at 17:20
• From your system, at first $$x_1$$, then $$x_3$$ and we get an equation only in $$x_2$$ including the parameters. Apr 28, 2019 at 18:21