0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ you can only if it's known that $a \ne 0$, otherwise still do it but consider potential error going that route $\endgroup$ – user29418 Apr 27 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 $\endgroup$ – Lubbi Apr 27 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. $\endgroup$ – user376343 Apr 27 at 13:37
1
$\begingroup$

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 ...

$\endgroup$
  • $\begingroup$ 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 $$ \left(\begin{array}{ccc|c} 0 & 3-a & -2 & 2 - a \\ 1 & a & 2 & a \\ 0 & 1 & 0 & \frac{-a^2}{2-a^2} \end{array}\right) $$ $\endgroup$ – Lubbi Apr 27 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!) $\endgroup$ – Lubbi Apr 27 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 $x_1$ $\endgroup$ – Ross Millikan Apr 28 at 1:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.