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I will post the original question that was done as an example in a linear algebra lecture followed up by my confusions / additional questions :)

Original Question:

What is the span inside $\mathbb{R}^3$ of

$$\underline{u_1} = \begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}, \underline{u_2} = \begin{pmatrix} -1\\ 0\\ 2 \end{pmatrix}, \underline{u_3} = \begin{pmatrix} 0\\ 1\\ 3 \end{pmatrix}, \underline{u_4} = \begin{pmatrix} -1\\ 1\\ 5 \end{pmatrix}$$

We want to find vectors $\underline{b}$ such that:

$$ \underline{b} = \lambda_1 \underline{u_1} + \lambda_2 \underline{u_2} + \lambda_3 \underline{u_3} + \lambda_4 \underline{u_4} = A \underline{\lambda} $$

So we write:

$$ \left[ \begin{array}{cccc|c} 1&-1&0&-1&b_1\\ 1&0&1&1&b_2\\ 1&2&3&5&b_3 \end{array} \right] $$

Through elementary row operations we reach:

$$ \left[ \begin{array}{cccc|c} 1&0&1&1&b_2\\ 0&1&1&2&b_2-b_1\\ 0&0&0&0&2b_1-3b_2+b_3 \end{array} \right] $$

Thus it is consistent when $2b_1 -3b_2 + b_3 = 0$ which is the equation of a plane. So the span of the 4 vectors is this plane.

My question:

Why, once we reach RREF of the augmented matrix are we able to to just ignore the other two equations. Sure the row of zeroes if definitely more interesting but how do we know that the other two equations don't restrict the span at all?

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    $\begingroup$ It is because, regardless of the $b_1, b_2$ values on the previous two rows, we can always find $\lambda_1, \lambda_2, \lambda_3, \lambda_4$ that satisfy the requirements. Namely: $\lambda_1 = b_2$, $\lambda_2 = b_2-b_1$, $\lambda_3=\lambda_4=0$. So for any $(b_1, b_2, b_3) \in \mathbb{R}^3$ that satisfies $2b_1 -3b_2+b_3=0$, we can find $\lambda$ so that $A\lambda = (b_1,b_2,b_3)$. $\endgroup$
    – Michael
    Nov 21 '19 at 21:18
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    $\begingroup$ @Retsek for me it's unclear what does $b_i$ stands for and why $2b_1-3b_2+b_3=0$ is a plane. $\endgroup$ Nov 21 '19 at 21:50
  • $\begingroup$ @Alexey I'm going to be completely honest I don't completely understand myself, these are just my notes from the lecture. It looks like 2x - 3y + z = 0 which is a plane in 3-dimensions. $\endgroup$
    – Retsek
    Nov 21 '19 at 21:55
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Suppose $2b_1-3b_2+b_3=0$. Then whatever values you choose for $b_1$ and $b_2$, the linear system is solvable. The two equations add no relevant information.

However you can use the RREF of the matrix formed with the four vectors to draw other conclusions. For instance that $u_1$ and $u_2$ form a basis of the span and that $$ u_3=u_1+u_2,\qquad u_4=u_1+2u_2 $$ because elementary row operations don't change linear relations between columns. So, since clearly the third column of the RREF is the sum of the first two, also the third column in the original matrix is the sum of the first two. Similarly for the fourth column. Since the first two columns in the RREF are linearly independent and form a basis of the span of the RREF's columns, the first two columns of the original matrix are a basis of the span of the four columns.

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