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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?

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    You can cancel the $F_i$'s.2011-04-15
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    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...)2011-04-15
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    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).2011-04-15
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    You can also cancel the $N$ if you want. Solve if for RHS which is constant $1$, then multiply the result by $N$.2011-04-15
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    There's a typo above; it should read $\sum_j A_{ij} x_j = B_i$.2011-04-15
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    @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

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