1
$\begingroup$

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

$B_{i}=F_i*N$

$A_{i,j}=-F_i/K_{j,i}$

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

$F_i=Y_i/K_{i,i}$

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?

  • 0
    @YuvalFilmus Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2015-05-05

1 Answers 1

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 $N$.