So say I have some linear transformation $T(\vec{x}) = A\vec{x}$

So we say $$ \left[ \begin{array}{ccc} a_1 & \dots & a_n\\ a_1 & \ddots & \vdots\\ a_1 & \dots & a_n\\ \end{array} \right] \times \left[ \begin{array}{c} x_1 \\ \vdots \\ x_n \\ \end{array} \right] = \left[ \begin{array}{c} b_1 \\ \vdots \\ b_n \\ \end{array} \right] $$

So my question is, if we take the following augmented matrix: $$ \left[ \begin{array}{ccc|c} a_1 & \dots & a_n & b_1\\ a_1 & \ddots & \vdots & \vdots\\ a_1 & \dots & a_n & b_n\\ \end{array} \right] $$ If we were to put the above matrix into reduced row echelon form, (assuming we are given some numbers to plug into matrix A), while leaving the b vector in terms of $b_1, b_2, ... b_n$, the right side of the matrix would end up being just a bunch of combinations of the terms $b_1, b_2, ... b_n$.

My question is, what does this resulting vector tell us about the range of this function.

I am interested in any usefulness that this new vector has, or any information we can glean from it.

  • $\begingroup$ yes, sorry about that. $\endgroup$ Sep 11, 2011 at 22:35

1 Answer 1


The question of whether $A\mathbf{x}=\mathbf{b}$ has a solution is equivalent to asking whether $\mathbf{b}$ lies in the columnspace of $A$.

Elementary row operations do not respect the columns space; however, every time you row-reduce the augmented matrix, you have a new system $A'\mathbf{x}=\mathbf{b}'$, and what you know is that $A\mathbf{x}=\mathbf{b}$ has a solution if and only if $A'\mathbf{x}=\mathbf{b}'$ has a solution, and in fact it has the same solutions. That means that if $\mathbf{b'}$ is in the column space of $A$, and you can write $$\mathbf{b'} = \alpha_1 A'_1 + \cdots + \alpha_k A'_k,$$ (where $A'_i$ is the $i$th column of $A'$, and $\alpha_i$ are scalars), then you will also have $$\mathbf{b} = \alpha_1A_1 + \cdots + \alpha_kA_k$$ (where $A_i$ is the $i$th column of $A$), with the same scalars. And conversely.

The point of going all the way to the reduced row-echelon form is that it makes it very easy to see what the solutions to $A'\mathbf{x}=\mathbf{b}'$ are; using those solutions you can express the original $\mathbf{b}$ as a linear combination of the columns of the original $A$.

(What is happening "behind the scenes" is that each time you do an elementary row operation, you are "really" performing an invertible linear transformation $E$, so that you are going from asking "is $\mathbf{b}$ in the range of $A$?" to asking "is $E\mathbf{b}$ in the range of $EA$?"; because $E$ is invertible, you can go back and forth from one question to the other).

(I'm not 100% sure that I am answering what you are asking; I hope I'm not too far off the mark)

  • $\begingroup$ This does answer my question, I'm currently trying to wrap my head around these ideas by playing around with different numbers. In the second paragraph, did you mean to say "and you know that A x = b has a solution if and only if A' x = b ' has a solution" $\endgroup$ Sep 11, 2011 at 23:18
  • $\begingroup$ @weezybizzle: Yes, I did; thank you. I'll correct. $\endgroup$ Sep 11, 2011 at 23:19

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