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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.

END

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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2  
I'm afraid this makes little sense... –  David Mitra Dec 20 '12 at 2:13
    
Did you write this correctly, it doesn't make sense as written? –  Amzoti Dec 20 '12 at 2:18
    
Why is that? It is two equations which can be solved recursively written in a matrix. Why doesn't it make any sense? –  Frederik Brinck Jensen Dec 20 '12 at 2:35
    
@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(!))? –  Billy Dec 20 '12 at 2:57
    
@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=0.004 –  Frederik Brinck Jensen Dec 20 '12 at 3:01

1 Answer 1

up vote 1 down vote accepted

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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1  
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 $-0.004-DW+0.01=0$. –  neuguy Dec 20 '12 at 3:35

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