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There are 3 tanks filled to capacity with fresh water, all with a 100 liter capacity. At t=0, brine with .5 kg/l salt concentration flows into tank 1 at a 3 l/min rate. The other flows are: tank 1 -> tank 2 at 4 l/min,
tank 1 -> tank 3 at 3 l/min,
tank 2 -> tank 1 at 2 l/min,
tank 2 -> tank 3 at 2 l/min,
tank 3 -> tank 1 at 2 l/min,
tank 3 -> tank 2 at 0 l/min,
and excess water leaves tank 3 -> ocean at 3 l/min.

I am supposed to write the system of equations in matrix form and I have figured it all out except how to place the constant brine flow into the matrix format. So far my matrix looks like this:

$\begin {bmatrix}-.07&.02&.02\\ .04&-.04&0\\ .03&.02&-.05\end {bmatrix}$ $\begin {bmatrix}x1\\ x2\\ x3\end {bmatrix}$

Any help on placing that constant brine flow is much appreciated.

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You can add a fourth line to your matrix, representing "tank 4" which is really the outside world. Why are the values in your matrix so small instead of being the single digit whole numbers of the problem?

Added: OK, I see that. So presumably you start with pure water in the tanks and are trying to calculate the salt concentration as a function of time. Now the matrix multiply is making sense-let $x_1,x_2,x_3$ be kg of salt in each tank. One equation would be $x_1'=1.5-0.07x_1+0.02x_2+0.02x_3$ so putting the outflows on the diagonal is correct. You should just add a column vector which is the input from the outside world, getting $$\begin {bmatrix} x_1'\\x_2'\\x_3'\end {bmatrix}= \begin {bmatrix}-.07&.02&.02\\ .04&-.04&0\\ .03&.02&-.05\end {bmatrix}\begin {bmatrix}x1\\ x2\\ x3\end {bmatrix}+\begin {bmatrix}1.5\\ 0\\ 0\end {bmatrix}$$

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The numbers are that way because it is x kg/l * l/min so, for example, you get (x kg/ 100 l) * (3 l/ 1 min) so 3/100 = .03. Also, wouldn't adding a fourth line just cause us to change the constant to a variable since your vertical x matrix would need another x4 or x0 added to it? – Gerard Apr 29 '13 at 17:44
@Gerard: The matrix still doesn't look right. Where did the diagonal elements come from? Yes, you would expand the matrix to $4 \times 4$ and make a new variable $x4$, which is the amount of water in the outside world. You can just ignore it at the end, but you need it for flow in/out. – Ross Millikan Apr 29 '13 at 18:05
The diagonal elements come from the rate out since the equation I am using is x = rate in - rate out. Is that not the way that you would solve a problem such as this? – Gerard Apr 29 '13 at 18:13
@Gerard: The matrix seems a reasonable way to show the flows, but I don't see how a matrix multiplication will help. Presumably you want to get something like $x_1'=+.03-.04-.03+.02+.02=0$ so the level in tank 1 is constant. – Ross Millikan Apr 29 '13 at 18:21
If you draw a diagram then you will see that the levels in each of the tanks stays constant at 100 liters. – Gerard Apr 29 '13 at 18:24

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