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I am attempting to find a 'smarter' way to solve a matrix, in the form $Ax=B$, where



where $N$ is constant, $K$ is a constant matrix, and $F$ is a vector of;


where $Y$ is a vector constant.

I'm not hugely mathematical, and my linear algebra skills could be put on the back of a napkin, but seeing this kind of repetition indicates to me that there must be a shortcut for this.

Does anyone have any insights?

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You can cancel the $F_i$'s. – Yuval Filmus Apr 15 '11 at 1:20
I was thinking more about the structure of the resultant matrix but yeah, I wanted a second opinion on that too (I have no real understanding of the weirdnesses of linalg...) – Andrew Bolster Apr 15 '11 at 1:23
Think of it as a system of equations. For each $i$ you have an equation of the form $\sum_j A_{ij} x_i = B_i$. In your case, this is $\sum_j -F_i/K_{ji} = NF_i$, so you can cancel the $F_i$'s. Otherwise it's almost completely arbitrary (other than the fact that $-1/K_{ji}$ cannot be zero). – Yuval Filmus Apr 15 '11 at 1:44
You can also cancel the $N$ if you want. Solve if for RHS which is constant $1$, then multiply the result by $N$. – Yuval Filmus Apr 15 '11 at 1:45
There's a typo above; it should read $\sum_j A_{ij} x_j = B_i$. – joriki Apr 15 '11 at 7:50

1 Answer 1

Let's write your system of equations in full. For each $i$, you have an equation $$ \sum_j (-F_i/K_{ij}) x_j = F_i N. $$ As you can see, you can cancel the $F_i$ (assuming for simplicity that they are non-zero): $$ \sum_j (-1/K_{ij}) x_j = N. $$ Other than that, the matrix $A$ is rather arbitrary (aside for having no non-zero entries).

Another simplification is to get rid of $N$: you can solve the system for $N = 1$, then multiply the solution by $N$.

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