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$$(a+1)x+y+3z=1$$$$8x+2y+(a+3)z=2$$$$3x+y+2z=-1$$

This question can be calculated with Gauss-elimination and I want to take away the y by taking $-2$ from the middle and $-1$ from the top and keep the third one as it is, my question is how do I calculate it with a parameter?

Inceptio
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user1838781
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  • (2a+2)x+2y+6z=2? – long tom May 14 '13 at 08:22
  • @long tom No, that's wrong – user1838781 May 14 '13 at 08:27
  • @user1838781 Parameter is just a fixed number. So, best way to do it, is simply ignore its unknown nature and operate as if you know it. And btw, long tom gave you quite correct hint. If you subtract expression he provided from your second equation, you'll get another expression without $y$ in it. – Kaster May 14 '13 at 08:36
  • @user1838781 i just multiplied the first equation by 2 to show how to handle parameters... – long tom May 14 '13 at 08:45
  • @Kaster that didn't help, I just want someone to describe with figures how I calculate with Gauss elimination when I have a parameter and your explanation didn't help – user1838781 May 14 '13 at 08:48
  • Unless you are interested in integer solutions only, [tag:linear-algebra] would be a much better tag than [tag:diophantine-equations]. – Martin Sleziak May 14 '13 at 13:07

1 Answers1

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$$ \left \{ \begin{array}{lcr} (a+1)x+y+3z & = & 1 \\ 8x+2y+(a+3)z & = & 2 \\ 3x+y+2z & = & -1 \end{array}\right . \\ \left \{ \begin{array}{lcr} 2(a+1)x+2y+6z & = & 2 \\ 8x+2y+(a+3)z & = & 2 \\ 3x+y+2z & = & -1 \end{array}\right . \\ \left \{ \begin{array}{lcr} 2(a-3)x+(3-a)z & = & 0 \\ 8x+2y+(a+3)z & = & 2 \\ 3x+y+2z & = & -1 \end{array}\right . \\ \left \{ \begin{array}{lcr} (a-3)(2x-z) & = & 0 \\ 8x+2y+(a+3)z & = & 2 \\ 3x+y+2z & = & -1 \end{array}\right . $$ So, from here, you can conclude that if $a = 3$, the system gets overdefined (number of variables exceeds number of equations), and since first equation will give you nothing, second and third equations can be restated taking into account $a = 3$ $$ \left \{ \begin{array}{lcr} 8x+2y+6z & = & 2 \\ 3x+y+2z & = & -1 \end{array}\right . $$ Subtract, doubled second from first $$ 2x+2z =4 $$ or $z = 2 - x$, substitute it to third equation back, so you'll get $$ y = -1 - 3x - 4+2x = -x-5 $$ As you may see, due to the overdefinition, you observed one possible scenario – parametric solution, i.e. when one of the variables can take any value, uniquely defining other two.

Now, let's assume that $a \ne 3$. In this case initial system can be restated as $$ \left \{ \begin{array}{lcr} 2x-z & = & 0 \\ 8x+2y+(a+3)z & = & 2 \\ 3x+y+2z & = & -1 \end{array}\right . $$ First equation yields $z = 2x$. Consider latter two $$ \left \{ \begin{array}{lcr} 8x+2y+2(a+3)x & = & 2 \\ 3x+y+4x & = & -1 \end{array}\right . \\ \left \{ \begin{array}{lcr} 2(a+7)x+2y & = & 2 \\ 7x+y & = & -1 \end{array}\right . \\ \left \{ \begin{array}{lcr} (a+7)x+y & = & 1 \\ 7x+y & = & -1 \end{array}\right . $$ Consider first equation minus second $$ ax+7x+y-7x-y = ax = 2 $$ Again, you can see that $a \ne 0$. If it is, first and second equations are inconsistent, so you won't get any solution from there.

And finally, you can find that $$ x= \frac 2a $$ Substitute it to the equations $$ y = -1 - 7x = -1 - \frac {14}a = -\frac {a+14}a $$ and recalling initial equation $z = 2x$ you can finally find that $$ \left [ \begin{array}{lc} a \notin \{0,3\} & \left \{ \begin{array}{ccr} x & = & \frac 2a \\ y & = & -\frac {a+14}a \\ z & = & \frac 4a \end{array} \right . \\ a = 3 & \left \{ \begin{array}{ccr} y & = & -x-5 \\ z & = & 2-x \end{array} \right . \end{array} \right . $$

Disclaimer

Even though this solution is overly filled with elementary things that could and even should have been easily omitted, I've solely decided that it might be useful for you. You can check answers provided here with Wolfram alpha by this link.

Kaster
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