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Introduction

Gaussian elimination is an algorithm used to solve system of equations. In the physical world, this technique can be used it to balance chemical equations, analyze electrical networks, and study the flow of traffic at a particular intersection.

Balancing Chemical Equations

An application of Gaussian elimination comes in the form of balancing equations which represent a chemical reaction. By definition, a chemical reactions occurs when a number of molecules combine to form new molecules.

In general, balancing a chemical reaction using matrices requires us to first, designate variables for the reactants and products. Secondly, we then equate the various elements from the left hand of the equation to the right hand components; this will generate a set of equations. To solve this set of equations, we then need to convert it into a homogeneous system, and represent it as a coefficient matrix. The rows of the matrix represent the quantity of each element in the unbalanced equations, whereas the columns represent the variables used to designate the number of molecules that are required by each compound in the final balanced equation.

To determine the balanced form of the equation, Gaussian elimination will be applied onto the matrix until a reduced row echelon form has been attained. Generally, a parameter will be required. The value of this parameter will be the smallest number which forces all other values to be positive and whole, since all reactions require integral molecules as opposed to fractions of them to complete the reaction.

Example

Photosynthesis is a process which harnesses the sun's energy, in the form of light, to convert carbon dioxide into sugar. Below is the chemical reaction for photosynthesis, which combines carbon dioxide with water to produce glucose and oxygen.

   $CO_{2}+H_{2}0\rightarrow C_{6}H_{12}O_{6}+O_{2}$

Balance this chemical equation.

Solution

The first step in balancing the reaction, is to designate the variables ${\it {{w},{\it {{x},{\it {{y},{\it {{z}\in N}}}}}}}}$ such that

   ${\it {{w}CO_{2}+{\it {{x}H_{2}0\rightarrow {\it {{y}C_{6}H_{12}O_{6}+{\it {{z}O_{2}}}}}}}}}$

We now equate the number of carbon, hydrogen, and oxygen atoms to form the following set of equations:

   ${\begin{aligned}w&=6y\\2w+x&=6y+2z\\2x&=12y\\\end{aligned}}$

To simplify the process of obtaining the solution, the above set of equations has been converted into a homogeneous system:

   ${\begin{aligned}w-6y&=0\\2w+x-6y-2z&=0\\2x-12y&=0\\\end{aligned}}$

Remark: In both systems (preliminary and homogeneous) each line (equation) represents one of the elements in the reaction.

To solve the homogeneous system, we apply Gaussian Elimination onto the system until a reduced row echelon form has been attained.

   ${\begin{pmatrix}1&0&-6&0&0\\2&1&-6&-2&0\\0&2&-12&0&0\end{pmatrix}}$

What follows is a series of row operations that take the above matrix to RREF.

   ${\begin{pmatrix}1&0&-6&0&0\\2&1&-6&-2&0\\0&2&-12&0&0\end{pmatrix}}$

   $-2R_{1}+R_{2}\rightarrow R_{2}{\begin{pmatrix}1&0&-6&0&0\\0&1&6&-2&0\\0&2&-12&0&0\end{pmatrix}}$

   $-2R_{2}+R_{3}\rightarrow R_{3}{\begin{pmatrix}1&0&-6&0&0\\0&1&6&-2&0\\0&0&-24&4&0\end{pmatrix}}$

        ${\frac {-1}{24}}R_{3}\rightarrow R_{3}{\begin{pmatrix}1&0&-6&0&0\\0&1&6&-2&0\\0&0&1&{\frac {-1}{6}}&0\end{pmatrix}}$

   $-6R_{3}+R_{2}\rightarrow R_{2}{\begin{pmatrix}1&0&-6&0&0\\0&1&0&-1&0\\0&0&1&{\frac {-1}{6}}&0\end{pmatrix}}$

     $6R_{3}+R_{1}\rightarrow R_{1}{\begin{pmatrix}1&0&0&-1&0\\0&1&0&-1&0\\0&0&1&{\frac {-1}{6}}&0\end{pmatrix}}$

Observe that we have four variables, but only three equations. As a result, we need to set a parameter. So from our final matrix, we see that

  ${\begin{aligned}w-z&=0\\x-z&=0\\y-{\frac {1}{6}}z&=0\\\end{aligned}}$

Which is equivalent to

  ${\begin{aligned}w&=z\\x&=z\\y&={\frac {1}{6}}z\\\end{aligned}}$

If we let $z=t$, then

  ${\begin{aligned}w&=t\\x&=t\\y&={\frac {1}{6}}t\\z&=t\\\end{aligned}}$

Smallest value for which $z$ can be set to so that we have postive and whole numbers for the other variables is $6$. So $w=6,x=6,y=1$, and $z=6$. The balanced equation is therefore:

   $6CO_{2}+6H_{2}O\rightarrow C_{6}H_{12}O_{6}+6O_{2}$