1
$\begingroup$

There are a set of equations like

$A_x + A_y + A_z = P$
$B_x + B_y + B_z = Q$
$C_x + C_y + C_z = R$

Where the values of only $P, Q, R$ are known.

Also, we have

$A_x + B_x + C_x = I$
$A_y + B_y + C_y = J$
$A_z + B_z + C_z = K$

where only the values of $I, J$ and $K$ are known.

Is there any way we know the individual values of

$A_x, B_x, C_x, A_y, A_z$ and the rest?

Substituting the above equations yield the result that $I + J + K = P + Q + R$ but how can I get the individual component values? Is any other information required to solve these equations?


Here's a good complementary question. If solutions exist, how to generate all of them? Are there some algorithms?

$\endgroup$
  • 1
    $\begingroup$ I suppose you mean $P+Q+R=I+J+K$ (there are no $X,Y,Z$ in the equations). This is indeed a necessary condition for the existence of a solution. Supposing that, you've got effectively $5$ linear equations (since one is dependent on the others) in $9$ unknowns; you cannot hope for a unique solution. $\endgroup$ – Marc van Leeuwen Oct 31 '13 at 9:42
  • $\begingroup$ @MarcvanLeeuwen Thanks i corrected it I +J + K = P +Q +R, Also, just now a friend told that there should be 9 equations to solve and get 9 unknow's. I guess that settle's it. $\endgroup$ – Neil Oct 31 '13 at 9:51
1
$\begingroup$

The matrix $(\begin{smallmatrix}+1&-1\\-1&+1\end{smallmatrix})$ has all row and column sums zero. So given any matrix$~A$ with at least two rows and at least two columns, you can always add a multiple of this matrix to a $2\times2$ submatrix of $A$ to obtain a different matrix $A'$ with the same row and column sums. So for such matrices, the row and column sums never determine the matrix.

$\endgroup$
  • $\begingroup$ I think we need additional information to solve these equations. I will check. $\endgroup$ – Neil Nov 1 '13 at 9:10
0
$\begingroup$

You are trying to solve $$\left(\begin{matrix} 1&1&1&0&0&0&0&0&0\\ 0&0&0&1&1&1&0&0&0\\ 0&0&0&0&0&0&1&1&1\\ 1&0&0&1&0&0&1&0&0\\ 0&1&0&0&1&0&0&1&0\\ 0&0&1&0&0&1&0&0&1 \end{matrix}\right) \left(\begin{matrix} A_x\\A_y\\A_z\\B_x\\B_y\\B_z\\C_x\\C_y\\C_z \end{matrix}\right) = \left(\begin{matrix} P\\Q\\R\\I\\J\\K \end{matrix}\right)$$ If you have any solution, it will be at least a four*-dimensional solution space, and to have a solution, necessarily the equation $P+Q+R=I+J+K$ must hold, as you already found out.

*If the LHS is $Ax$ then the dimension (if $\ge 0$) is at least $9 - {\rm rg}(A) = 9-5 = 4$, the rank is only five because the sum of the first three rows is equal to the sum of the last three rows.

$\endgroup$
  • $\begingroup$ @MarcvanLeeuwen Thanks for pointing out that the Rank of the $6\times 9$ matrix is only $5$ ;) $\endgroup$ – AlexR Oct 31 '13 at 10:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.