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In the MIT lecture notes (pg. 1) for Linear Algebra, it states;

The third row is zero because row 3 was a linear combination of rows 1 and 2; it was eliminated.

Here, $A = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 8 & 10 \end{bmatrix}$ and $U = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 \end{bmatrix}$.

Does this mean that if the row of the echelon matrix is all zeros, the corresponding row in A can be represented as a linear combination of the other rows in A?

And if the above statements is true, does this also apply to reduced row echelon matrix?

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(1) Yes, if a row of the echelon matrix is all zeros, then that row (but remember to consider possible partial pivoting) is a linear combination of other rows in the original matrix. Generally, the coefficients such that forms this linear combination can be found solving the following linear system for $a$ and $b$: $$\begin{bmatrix}1 & 2 \\ 2 & 4 \\ 2 & 6 \\ 2 & 8 \end{bmatrix}\cdot\begin{bmatrix}a \\ b \end{bmatrix}=\begin{bmatrix}3 \\ 6 \\ 8 \\ 10\end{bmatrix}$$

(2) Yes, since it applies to echelon forms it also applies to row reduced echelon forms.

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