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I have the following equation system I need to convert into a reduced echelon form. I have issues with the unknown number, a.

  • $2x_1 + (3 + a)x_2 + 2x_3 = 2 + a$
  • $x_1 + ax_2 + 2x_3 = a$
  • $ax_1 + 2x_2 + 2ax_3 = 0$

First I convert the above linear system into a matrix:

$$ \left(\begin{array}{ccc|c} 2 & 3+a & 2 & 2 + a \\ 1 & a & 2 & a \\ a & 2 & 2a & 0 \end{array}\right) $$

Normally I can fairly easy use Gauss' forward and backward elimination to create zeros under and above each pivot. However, with $a$ in the picture I struggle to reduce it any further. For example if I were to reduce the last row, I believe I can't just say $a \times 1/a $ ? Since a could be $0$.

I'm not sure I can go about this. Hope someone can help me!

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  • $\begingroup$ you can only if it's known that $a \ne 0$, otherwise still do it but consider potential error going that route $\endgroup$ Commented Apr 27, 2019 at 10:45
  • $\begingroup$ Thank you for the answer. I've tried but am stuck trying to reduce the $a$ in the first column still. If I can't divide by $\frac{1}{a}$ then I don't know how to get $0$'s below the first pivot $\endgroup$ Commented Apr 27, 2019 at 11:19
  • $\begingroup$ @Lubbi you can do first your computation for $a=0.$ If I am right, the solution is $(2,0,-1).$ Then you can assume $a\neq 0.$ Do you know Cramer's rule (determinants)? For it, a condition $2a^2-4\neq0$ will appear. The rest is standard. $\endgroup$ Commented Apr 27, 2019 at 13:37

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You just do it normally, carrying along the $a$s as needed. Any time you need to divide by something, set that to zero and solve the system. The $a$s will be gone. If you want to divide by $a$, first consider what happens when $a=0$ $$\left(\begin{array}{ccc|c} 2 & 3+a & 2 & 2 + a \\ 1 & a & 2 & a \\ a & 2 & 2a & 0 \end{array}\right)=\left(\begin{array}{ccc|c} 2 & 3 & 2 & 2 \\ 1 & 0 & 2 & 0 \\ 0 & 2 & 0 & 0 \end{array}\right)$$ which is a form you know how to solve. I get $(2,0,-1)$ as a solution. Now you can assume $a\neq 0$ and divide by it. If I use the second row to clear the first column I get $$\left(\begin{array}{ccc|c} 0 & 3-a & -2 & 2 - a \\ 1 & a & 2 & a \\ 0 & 2-a^2 & 0 & -a^2 \end{array}\right)$$ Now we need to divide by $2-a^2$, so we have to worry about $a=\pm 2$, so check that, then divided the third row by it and use it to zero the second element in the first row. You can then report the solution as "If $ a= \sqrt2,$ such and such, if $a=-\sqrt 2,$ so and so, otherwise ...

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  • $\begingroup$ Thanks, Ross. But for $a \neq +-\sqrt 2,$, how would you show this in a reduced echelonform? I get some really crazy calculations with $a$ in the last column. I'm not sure I'm on the right track I land here and show that $$ \left(\begin{array}{ccc|c} 0 & 3-a & -2 & 2 - a \\ 1 & a & 2 & a \\ 0 & 1 & 0 & \frac{-a^2}{2-a^2} \end{array}\right) $$ $\endgroup$ Commented Apr 27, 2019 at 20:18
  • $\begingroup$ So I show atleast that $a \neq \pm \sqrt{2}$. But how do I reduce the matrix further, into a reduced echelonform? Sorry I've started learning about Linear Algebra last week. (And thank you, really. You helped me a lot further!) $\endgroup$ Commented Apr 27, 2019 at 20:25
  • $\begingroup$ You are almost there. Now subtract $3-a$ times the third row from the first. Then divide the first by $-2$. Put the second row first, the third second, and the first third and you are there. You can read off the values of $x_3$ and $x_2$, then backsubstitute to get $x_1$ $\endgroup$ Commented Apr 28, 2019 at 1:58

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