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Hi there math experts.

I have the following matrix:

$$ \begin{equation} \begin{pmatrix} -1x & 0y & 0 & 0 & 0.004 & 0\\ -1x & -1y & 0 & 0 & 0.001 & 0 \end{pmatrix}=0 \end{equation} $$

Elaborating on the question:

I'm currently finishing a paper in which I have to deal with a economic model. I try to give a very brief and simple example of how it works through matrix calculation. Based on two equations with two unknown variables (I'm aware that their actually aren't two, but only one, since the other can be determined recursively, though it still serves the purpose) I put up the following equation.

$$ \begin{align*} \begin{pmatrix} -1 & 0 & -1.407\cdot 0 & 0.0407\cdot 0 & 0.4\cdot 0.01 & -0.0337\cdot 0 \\ -1 & -1 & 0 \cdot 0 & 1\cdot 0 & 1\cdot 0.01 & 0\cdot 0\\ \end{pmatrix} \cdot \begin{pmatrix} DC\\ DW\\ 0 \\ 0 \\ 0.01 \\ 0 \end{pmatrix}=0 \end{align*} $$

Now I have to determine $DC$ and $DW$, which I do by multiplying the scalar with the matrix, which is why I get the initial matrix, as stated in the question above.

I know that the result should be $DC=0.4$ and $DY=0.6$ since I've been running some tests on the computer. I just can't figure it out in the hand though, and it is necessary for me to write the solution down step by step.

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Whenever I solve this by hand, I get $x =0.04$ and $y=-0.003$, though when I sovle it, for instance, in Mathematica, i the $y=0.006$ which is the correct result.

What am I overlooking?

Best regards. /brinck10

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    I'm afraid this makes little sense...2012-12-20
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    Did you write this correctly, it doesn't make sense as written?2012-12-20
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    Why is that? It is two equations which can be solved recursively written in a matrix. Why doesn't it make any sense?2012-12-20
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    @FrederikBrinckJensen Is your right hand side the *scalar* 0 (in which case you have matrix = scalar, which is nonsense), or the *matrix* 0 (in which case every entry is equal to 0, in particular x = 0, y = 0, and 0.004 = 0(!))?2012-12-20
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    @Billy the way I understood it was, that I could write the matrix M multiplied with the scalar K giving 0, so that M.K=0. In that case I would have a coefficient matrix, where I could just solve each line in regard to 0, i.e -1x + 0y + 0 + 0 + 0.004 + 0 = 0 <=> x=0.0042012-12-20
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    @FrederikBrinckJensen Given the question as stated, I see no reason for those plus signs to pop up. Perhaps you've written it wrongly. Can you go back a step or two and tell us how you got to this equation? Also, what makes you think y = 0.006 is correct?2012-12-20
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    Can anyone explain how to interpret this question?2012-12-20
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    @Billy, see my explanation!2012-12-20
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    @FrederikBrinckJensen, you've forgotten the second matrix that you've included in your elaboration in the original question. No wonder the original didn't make sense (how could a matrix with a nonzero entry be 0?).2012-12-20
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    I'm sorry, hope it makes sense now. Any clue on how to solve it?2012-12-20
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    Figured out what the question was. Probably should have edited the post as well, but I think the answer does the job well enough.2012-12-20
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    Is $DC$ the same thing as $x$? Are $DW$, $DY$, and $y$ all names for the same thing?2012-12-20

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Your matrix equation, written properly, is $$ \begin{align*} \begin{pmatrix} -1 & 0 & -1.407 & 0.0407 & 0.4 & -0.0337 \\ -1 & -1 & 0 & 1 & 1 & 0\\ \end{pmatrix} \cdot \begin{pmatrix} DC\\ DW\\ 0 \\ 0 \\ 0.01 \\ 0 \end{pmatrix}= \begin{pmatrix} 0\\ 0 \end{pmatrix} \end{align*}. $$ Multiplying out, we get the system of linear equations $$ -DC + 0.004 = 0\\ -DC - DW + 0.01=0 $$

Solving for $DC$, we just move $DC$ to the right to get $DC=0.004$. Plugging into the second equation, we have $-0.004-DW+0.01 = -DW+0.006 = 0$. So $DW=0.006$.

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    Much of the confusion with this problem was due to bad formatting. Next time, do NOT multiply in the coefficients of the right matrix into the left matrix, then multiply by the right matrix. (Compare my matrices to yours.) What you were overlooking is probably that your second equation of the system read $-0.004-DW+0.001=0$, instead of $-0.004-DW+0.01=0$.2012-12-20