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Consider the following equation:

$\left[\array{1 & 0.1353 & 1 \\0.3678 & 0.3678 & 1 \\ 0.1353 & 1 & 1 \\ 0.3678 & 0.3678 & 1}\right]\left[\array{w_1\\w_2\\w_3}\right]=\left[ \array{ 0\\1 \\ 0 \\ 1}\right]$

The $4$ by $3$ matrix on the LHS, is not a square matrix and has infinite left inverses, but I think there is only one solution for $[w_1 w_2 w_3]^T$ because the rows $2$ and $4$ of the matrix are the same and have similar values on the RHS, so in fact we have $3$ linear equations for $3$ unknowns, so we have only one solution for this equation.

Am I right or I have made a mistake?

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    $\begingroup$ You can always row reduce the matrix to see what equations you really have. $\endgroup$ – ja72 Dec 9 '13 at 18:48
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Yes, because rows two and four are the same on both sides, you can simplify your system to $$ \begin{pmatrix}1 & .1353 & 1\\.3678 & .3678 & 1 \\ .1353 & 1 &1 \end{pmatrix} \begin{pmatrix} w_1 \\ w_2\\ w_3 \end{pmatrix}=\begin{pmatrix} 0\\1\\0 \end{pmatrix} $$ If the $3\times 3$ matrix has nonzero determinant, then you have a unique solution.

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