I am given three equations with two unknowns. I am asked to find an equation which must be satisfied by $a$, $b$, & $c$ in order for this system to have a solution.

$$ x - y = a \\ -2x + 3y = b \\ -x - 2y = c $$

I have used Reduced Row Echelon Form to solve for the pivot variables $x$ and $y$.

$$ x = 3a + b \\ y = 2a + b $$

At this point I am able to arbitrarily pick $a$ and $b$ with success for the first two equations but am confused with the third equation and $c$. At this point I have reasoned $c = -x - 2y = -7a - 3b$ after substituting the values found with rref. Solving in terms of $a$, $b$, and $c$, I finally arrive at $$-7a - 3b - c = 0$$

Logically this would be an equation that if satisfied would give the original system a solution. Is this the correct solution as asked? Is the logic used to derive it correct as well.

  • $\begingroup$ You went too far. $c=-7a-3b$ is the requested compatibility condition. But what do you mean by "Solving in terms of a , b , and c" ? $\endgroup$ – Yves Daoust Sep 12 '14 at 21:40
  • $\begingroup$ I mean that given any a, b, and c they relationship must equal zero. I thought the problem addressed whether any a, b, c would be a solution, not find a c that gives a solution for a and b. "In terms" identifies that a,b,c are given. $\endgroup$ – Ben Campbell Sep 12 '14 at 21:47
  • $\begingroup$ Sorry, this is not clearer, I can't figure how you arrive at $-7-3b-c=0$; but this is most probably just a typo when rewriting $c=-7a-3b$. $\endgroup$ – Yves Daoust Sep 12 '14 at 21:50
  • $\begingroup$ You are correct, it is a typo. $c = -7a - 3b$ is equivlent to $0 = -7a - 3b - c$ $\endgroup$ – Ben Campbell Sep 12 '14 at 21:58

Your solution is fine.

An alternate way to do this would be to reduce the augmented matrix to row echelon form to get

$\begin{bmatrix} 1&-1&a\\-2&3&b\\-1&-2&c\end{bmatrix}\longrightarrow\begin{bmatrix}1&-1&a\\0&1&b+2a\\0&-3&a+c\end{bmatrix}\longrightarrow\begin{bmatrix}1&-1&a\\0&1&b+2a\\0&0&7a+3b+c\end{bmatrix}$,

and then the system has a solution iff $7a+3b+c=0$.

  • $\begingroup$ A much less verbose answer. It is interesting to see the relationship expressed through different mechanisms. $\endgroup$ – Ben Campbell Sep 12 '14 at 22:22

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