Question 842463
Best I can figure from this little beauty is that there is no unique solution.
I solved it manually and got a row of all zeroes.
I then double checked what I did using the following matrix calculator.
<a href = "http://karlscalculus.org/cgi-bin/linear.pl" target = "_blank">http://karlscalculus.org/cgi-bin/linear.pl</a>
The calculator confirmed what I did manually.
It also wound up with a row of all zeroes.
The implication here is that there is no unique solution, and that any solution depends on the value of one of the number of bacterium in one of the species.


To solve this problem using a matrix, I set up the following consumption table:
<pre>
                species 1     species 2     species 3        total
resource 1          6             1             3            27000
resource 2          0             3             9             9000
resource 3          1             3             9            13000

</pre>


let:
a = number of bacterium of species 1
b = number of bacterium of species 2
c = number of bacterium of species 3


the equations that need to be solve simultaneously so you can get a unique solution of a, b, and c are:


6a + 1b + 3c = 27000
0a + 3b + 9c = 9000
1a + 3b + 9c = 13000


the augmented matrix that is set up to solve these equations simultaneously is shown below:


<pre>                
         6             1             3            27000
         0             3             9             9000
         1             3             9            13000
</pre>


this matrix is composed of the coefficients and constants of the equations to be solved.


I'll spare you the details of how I solved it manually.
I solved it both ways.
One way was using the matrix.
The other way was solving the equations directly.
Both methods yielded the same solution.
The third row became all 0's.
This means that there were solutions to the set of equations but there was no unique solution to the set of equations.
The phrase was that there were dependent solutions.


I then used one of the equations to solve for b in terms of c.
I then assumed values for c >= 0 and <= 1000.
c had to be less than or equal to 1000 or b would be negative which is not allowed.


I could make any value of c that I liked that was less than or equal to 1000 and I got a common solution for b and for a.


I'll show you a couple so you'll get the idea.


The equations that needed to be solved simultaneously are:


6a + 1b + 3c = 27000
0a + 3b + 9c = 9000
1a + 3b + 9c = 13000


If I use the second equation to solve for b in terms of c, I get:
b = (9000 - 9c) / 3 which simplifies to:
b = 3000 - 3c


If I allow c to be equal to 1000, then b is equal to 3000 - 3000 = 0.
so, when c = 1000, b = 0 and I can solve for a in either equation that contains a.


The first equation is 6a + 1b + 3c = 27000
substituting for b and c, I get:
6a + 0 + 3000 = 27000
solving for a, I get a = 4000


the third equation is a + 3b + 9c = 13000
My common solutions are a = 4000 and b = 0 and c = 1000
substituting in this equation, I get:
4000 + 0 + 9000 = 13000.
The values of a, b, and c are good.


I have a common solution for all 3 equations and that common solution is:
a = 4000
b = 0
c = 1000


Now assume c is equal to 500.
It's less than 1000 so it meets the requirement for c to be greater than or equal to 0 and less than or equal to 1000.


The value of b is dependent on c.
The equation is b = 3000 - 3c
When c = 500, this equation becomes b = 3000 - 1500 which makes b = 1500
We have:
c = 500
b = 1500.


If we substitute in the third equation this time (just to make it different - didn't have to do it this way), then we get:
a + 3b + 9c = 13000
substituting for b and c, we get:
a + 4500 + 4500 = 13000
This simplifies to:
a + 9000 = 13000
Solving for a gets:
a = 4000.


It appears that a may have solidified to the value of 4000.
substituting a = 4000 and b = 1500 and c = 500 in the first equation gets:
6a + 1b + 3c = 27000 which becomes:
24000 + 1500 + 1500 = 27000 which becomes:
27000 = 27000 which confirms the solution of a = 4000 and b = 1500 and c = 500 as good.


We now have 2 solutions that satisfy the equation.


We'll do one more just to see if the value of a changes.


Let's  allow c to be equal to 100.
b is equal to 3000 - 3c which makes b = 3000 - 300 = 2700.


we have:
b = 2700
c = 100


let's go to the first equation and solve for a.
The first equation is:
6a + 1b + 3c = 27000
This becomes:
6a + 2700 + 300 = 27000
This results in:
6a + 3000 = 27000 which results in:
6a = 24000 which results in:
a = 4000


Looks like the value of a will always be 4000, while the value of b will depend on the value of c.


let's confirm by substituting in the last equation.
The last equation is:
1a + 3b + 9c = 13000
Substituting gets us:
4000 + 3*2700 + 9*100 = 13000
Simplifying this gets us:
4000 + 8100 + 900 = 13000
Simplifying further gets us:
4000 + 9000 = 13000 which is true.


We have common solutions but they are not unique.
This confirms what the matrix solution told us when we got a row of all zeroes.
i told us that we had a dependent solution that was not unique.


a is equal to 4000
the value of b is dependent on the value of c.


I'm assuming that if I solved for c in terms of b, then the dependency would have been that c was dependent on the value of b.


Either way we have a dependent solution.


The way we solved it, b is dependent on the value of c and the value of c has to be between 0 and 1000 in order to be valid.


There is another matrix calculator that I used to confirm the solution was as I had suspected.


That calculator does the same job as the calculator I referenced above, but it does it without the verbiage.


The first calculator is good for explaining what is going on.
The second calculator just does the calculations and shows you the result.


The second calculator can be found at this link:


<a href = "http://www.gregthatcher.com/Mathematics/GaussJordan.aspx" target = "_blank" >http://www.gregthatcher.com/Mathematics/GaussJordan.aspx</a>


They both perform the calculations using the gauss-jordan method of elimination.