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Solve $$\begin{matrix}i \\ ii \\ iii\end{matrix}\left\{\begin{matrix}x-y-az=1\\ -2x+2y-z=2\\ 2x+2y+bz=-2\end{matrix}\right.$$

For which $a$ does the equation have

  • no solution
  • one solution
  • $\infty$ solutions

I did one problem like this and got a fantastic solution from @amzoti. Now, I think that if I see another example, I will really get it.

EDIT

Here is my attempt with rref and here with equations

Problems

  1. I don't know how to handle the $b$ in the end.
  2. Does it ever lead to, speaking in "matrix terms", the case 0 0 0 | 0 so that I'll have $\infty$ number of solutions?
Community
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jacob
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  • I am afraid that you will have to do Gauss elimination method. Are you familiar with that? – imranfat Oct 31 '13 at 19:06
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    Either you understood Amzoti's solution and you can do the same here or at least show us how you tried to. Or you did not understand Amzoti's solution back there and you should ask for clarifications before asking another very similar question. – Julien Oct 31 '13 at 19:08
  • Even without Gauss elimination, what immediately leaps out is the coefficients of $x$ and $y$ in the first two equations: If you multiply the first equation by 2 and add it to the second, you get an equation in which neither $x$ nor $y$ appear. That should make the whole system easy to unravel. – Harald Hanche-Olsen Oct 31 '13 at 19:09
  • So Jacob, give it a try first and show us what you have, we can then help you to reach the solution. – imranfat Oct 31 '13 at 19:09
  • @HaraldHanche-Olsen. Yes, I realized that too after typing, I was just going very general here...... – imranfat Oct 31 '13 at 19:11

3 Answers3

1

The complete matrix of your system is $$ \begin{bmatrix} 1 & -1 & -a & 1 \\ -2 & 2 & -1 & 2\\ 2 & 2 & b & -2 \end{bmatrix} $$ and with Gaussian elimination you get $$ \begin{bmatrix} 1 & -1 & -a & 1 \\ 0 & 0 & -1-2a & 4\\ 0 & 4 & b+2a & -4 \end{bmatrix} $$ (sum to the second row the first multiplied by $2$ and to the third row the first multiplied by $-2$). Now swap the second and third rows: $$ \begin{bmatrix} 1 & -1 & -a & 1 \\ 0 & 4 & b+2a & -4 \\ 0 & 0 & -1-2a & 4 \end{bmatrix} $$ You see that you have solutions if and only if $-1-2a\ne0$. Otherwise the last equation would become $0=4$ that obviously has no solution.

The solution is unique for $a\ne-1/2$ and does not exist for $a=-1/2$.

egreg
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  • Did I calculate it correctly? 2) I thought you had to have 1 on the left side of your | in the augmented matrix, no? 3) We don't care about b?
    • – jacob Oct 31 '13 at 22:10
  • @jacob If it's just to know the number of solutions, reducing the pivots to $1$ is not important; if you have to tell what the solution is, then it's better to do it. This system can't be indeterminate for any value of $a$ and $b$. Of course the solution will depend on $a$ and $b$. – egreg Oct 31 '13 at 22:15