# Is it possible to solve for $x$ in this system?

Given: \begin{align*} a + 2b &= 5c\\ 2c + 2b &= 2a\\ a + 3c &= xb \end{align*}

Is it possible to solve for $x$?

I figure it's impossible, but I wanted to be sure before sticking my foot in my mouth (this is a silly bet in my office).

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Are $a$, $b$, and $c$ specific, given numbers? Or do you mean "can we express $x$ in terms of $a$, $b$ and $c$? – Arturo Magidin Feb 16 '12 at 17:37
In your title, you ask if it's possible to solve for (capital) $X$; then later, you ask whether it's possible to solve for (lower-case) $x$. – Michael Hardy Feb 16 '12 at 17:42
I mean, is it possible to find a specific value for x, not in terms of a, b, and c. – Jordan Hudson Feb 16 '12 at 17:47
@JordanHudson: Sorry, but that question makes no sense as written. Is it possible to find a specific value for $x$ that will satisfy what condition? Make the system consistent? Make the system have a unique answer? Make your significant others wiggle their toes? – Arturo Magidin Feb 16 '12 at 17:51

The system is equivalent to: $$\begin{array}{rcccccl} a & + & 2b & - & 5c & = & 0\\ -2a & + & 2b & + & 2c & = & 0\\ a & - & xb & + & 3c & = & 0 \end{array}$$ Performing Gaussian elimination, we have: \begin{align*} \left(\begin{array}{rrr} 1 & 2 & -5\\ -2 & 2 & 2\\ 1 & -x & 3 \end{array}\right) &\to \left(\begin{array}{rrr} 1 & 2& -5\\ 0 & 6 & -8\\ 0 & -x-2 & 8 \end{array}\right)\\ &\to \left(\begin{array}{rrr} 1 & 2 & -5\\ 0 & 6 & -8\\ 0 & 4-x & 0 \end{array}\right).\end{align*} If $x=4$, then the system has infinitely many solutions for $a$, $b$, and $c$, given by $a=\frac{7}{3}c$, $b=\frac{4}{3}c$.

If $x\neq 4$, then the only solution to the system is $a=b=c=0$.

So, in a sense, $x$ can be anything if $a$, $b$, and $c$ satisfy $a=\frac{7}{3}c$, $b=\frac{4}{3}c$, $c$ arbitrary; and if $a$, $b$, and $c$ do not satisfy those conditions, then no value of $x$ will give a consistent system.

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Perfect, thank you. – Jordan Hudson Feb 16 '12 at 17:49

When doing symbolic (vs. purely numeric) Gaussian elimination, it often helps to reorder the columns and rows so that the symbolic coefficients (here $x$) are on the lower right of the matrix. This helps the elimination to proceed purely numerically as far as possible. In particular, it will help the pivots to remain numeric for as long as possible, reducing complexity by reducing branching on whether or not a symbolic pivot is zero (necessary to avoid dividing by a zero).

In the example above, we can push the lone symbolic coefficient $x$ all the way down-right by simply swapping the last two columns, namely $$\left(\begin{array}{rrr} 1 & \!\!-5 & 2\\ -2 & 2 & 2\\ 1 & 3 & \!\!-x \end{array}\right)\ \to\ \left(\begin{array}{rrc} 1 & \!\!-5& 2\\ 0 & \!\!-8 & 6\\ \:0 & 8 & \!\!\!-x\!-\!2 \end{array}\right) \ \to\ \left(\begin{array}{rrc} 1 & \!\!-5 & 2\\ 0 & \!\!-8 & 6\\ \:0 & 0 & \!\!4\!-\!x \end{array}\right)$$

While this is a bit trivial above, in less trivial systems it can allow one to simply go much further towards full triangularization before having to grapple with complexities of symbolic elimination.

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