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Suppose I have a non-square matrix where some rows of values are known, and others are not.

\begin{bmatrix}x_{11}&x_{12}&x_{13}\\x_{21}&x_{22}&x_{23}\\ y_{31}&y_{32}&y_{33}\\x_{41}&x_{42}&x_{43}\\\end{bmatrix}

Assume that all X values are the known values (the entire rows' values will be known) and that there are rows where all the values are unknown, the Y values.

We also know a few other key points of information. We know the linear algebra equation comes out to XV = W where V is a vector of variables \begin{bmatrix}v_1\\v_2\\v_3 \end{bmatrix} and W is a vector that is the dot product of X and V. All values of W are known.

For example, the row of Y values, while we don't know the individual Y values, we know that $y_{31}*v_1 + y_{32}*v_2 + y_{33}*v_3 = w_3$ and we actually know the true value of $w_3$, just not the $y$ values that lead to it.

Is there a way to solve for possible values of the unknown $y$ values? Looking to understand the linear algebra behind the concept and will eventually imploy solution (if possible) in python.

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  • $\begingroup$ So you have a matrix with nine given entries, and you have one linear equation involving the other three entries? Then no, you can't determine all three of them from that. $\endgroup$
    – Ian
    Apr 22 '20 at 18:59
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You can't extract those values. Suppose $y_{31} v_1 + y_{32}v_2 + y_{33} v_3 = w_3$. Then $(y_{31} + v_2)v_1 + (y_{32} -v_2)v_1 + y_{33}v_3 = w_3$ also. So if at least two of your $v_j$ are nonzero, you will certainly not have a unique solution.

If you get to pick your vector $v$. Just take $v = (1,0,0)$, then $Xv = y_{31}$, $v=(0,1,0)$, gives you $Xv= y_{32}$ and so on.

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