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Three species of bacteria coexist in a test tube, feeding on three food sources. Suppose that a bacterium of species i consumes  
\[c_{ij}\]
  per day of food source j.
Let  
\[c_i=(c_{i1}, c_{i2}, c_{i3})\]
  be the consumption vector for species i. If  
\[c_1=(1,1,1), \; c_2=(1,2,3), \; c_3=(1,3,5)\]
  
\[\]
  and to start with there are 15,000, 30,000 and 45,000 units of food source 1, 2 and 3 respectively supplied each day, what are the population s of the three species that can coexist?
Assuming all the food is consumed, we can write down the equations:
\[x_1+x_2+x_3=15000\]

\[x_1+2x_2+3x_3=30000\]

\[x_1+3x_2+5x_3=45000\]

This is equivalent to the matrix system  
\[\left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{array} \right) \begin{pmatrix}x_1\\ x_2\\ x_3 \end{pmatrix} \begin{pmatrix}15000\\30000\\45000\end{pmatrix}\]
.
Form the augmented matrix  
\[\left( \begin{array}{cccc} 1 & 1 & 1 & 15000 \\ 1 & 2 & 3 & 30000\\ 1 & 3 & 5 & 45000 \end{array} \right) \]
.
Now row reduce.
R2-R1, R3-R1, :  
\[\left( \begin{array}{cccc} 1 & 1 & 1 & 15000 \\ 0 & 1 & 2 & 15000\\ 0 & 2 & 4 & 30000 \end{array} \right) \]
.
R1-R2, R3--2(R2), :  
\[\left( \begin{array}{cccc} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 15000\\ 0 & 0 & 0 & 0 \end{array} \right) \]
.
We can take  
\[0 \le x_3 \le 15000\]
  as arbitrary, then  
\[x_1=x_3, \; x_2=15000-2x_3\]