Balancing chemical equations using linear algebraic methods

I know there are already plenty of questions on this site regarding this topic but I am having difficulty with a particular chemical equation.

I am trying to balance the following:

$$ { C }_{ 2 }{ H }_{ 2 }{ Cl }_{ 4 }\quad +\quad { C }a{ { (OH }) }_{ 2 }\quad \xrightarrow [ ]{ } \quad { C }_{ 2 }{ H }{ Cl }_{ 3 }\quad +\quad Ca{ Cl }_{ 2 }\quad +\quad { H }_{ 2 }{ O } $$

The system of linear equations produces the following augmented matrix:

$$ \begin{pmatrix} 2 & 0 & -2 & 0 & 0 & 0 \\ 2 & 2 & -1 & 0 & -2 & 0 \\ 4 & 0 & -3 & -2 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 & -1 & 0 \end{pmatrix} $$

With the rows in the following order: Carbon Hydrogen Chlorine Calcium Oxygen

In row echelon form this reduces to:

$$ \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} $$

Which would indicate that: x1 = 0; x2 = 0; x3 = 0; x4 = 0; x5 = 0 which is obviously not correct. What have I done wrong?

Solution 1:

Let the balanced equation to be $$a\,{ C }_{ 2 }{ H }_{ 2 }{ Cl }_{ 4 }\quad +\quad b\,{ C }a{ { (OH }) }_{ 2 }\quad \xrightarrow [ ]{ } \quad c \,{ C }_{ 2 }{ H }{ Cl }_{ 3 }\quad +\quad d\,Ca{ Cl }_{ 2 }\quad +\quad e\ { H }_{ 2 }{ O }$$ parform the balance for each atom (in the order $C, H, Cl, Ca, O$). We then have $$2a=2c$$ $$2a+2b=c+2e$$ $$4a=3c+2d$$ $$b=d$$ $$2b=e$$ But one of the parameters must be fixed (say $a=1$). So, eliminating terms : $c=1$, $b=d=\frac e 2$, $d=\frac 12$, $e=1$ which make $a=c=1$, $b=d=\frac 12$, $e=1$. To get rid of the fractions, multiply all coefficients by $2$.

Solution 2:

The 4th row of your matrix is missing a minus sign, i.e. $$ \begin{pmatrix} 2 & 0 & -2 & 0 & 0 & 0 \\ 2 & 2 & -1 & 0 & -2 & 0 \\ 4 & 0 & -3 & -2 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & 0 \\ 0 & 2 & 0 & 0 & -1 & 0 \end{pmatrix} $$ $$ $$ But instead of forming an augmented matrix, you should drop that last column of zeroes, use the next-to-last column as a $b$-vector (equivalent to setting $x_5=1$), and solve for the remaining four coefficients from the linear system $$ A\,x = b $$ where $$ \eqalign{ A &= \begin{pmatrix} 2 & 0 & -2 & 0 \\ 2 & 2 & -1 & 0 \\ 4 & 0 & -3 & -2 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & 0 & 0 \end{pmatrix} \cr\cr b &= \begin{pmatrix} 0 \\ 2 \\ 0 \\ 0 \\ 1 \end{pmatrix} \cr }$$ This is an overdetermined system which can be solved as $$ \eqalign{ x &= (A^TA)^{-1}A^T\,b \cr &= A^+\,b \cr }$$ Finally, you can multiply the x-vector by a scalar factor to round any fractional components to whole integers.