Talk:Row Equivalent Matrix for Homogeneous System has same Solutions

I don't believe this page holds water as it stands.

Reference the first "explain" template specify exactly what you are multiplying by $a$: one row? One row of all the matrices? Because in the below you haven't multiplied the matrices, you've multiplied one of the equations. It needs to be shown that multiplying one row of $\mathbf A$ results in the change of the equation as specified.

And then, having multiplied that row of the equation by $a$, your array which includes $a \beta_i$ (where $a$ seems to have turned into $\lambda$) is no longer equal to $\mathbf b$.

So your initial statement:

$\mathbf A \sim \mathbf H \implies \left\{{\mathbf x: \mathbf A \mathbf x = \mathbf b}\right\} = \left\{{\mathbf x: \mathbf H \mathbf x = \mathbf b}\right\}$

may indeed even not be true. --prime mover 03:00, 29 March 2012 (EDT)

Eeek! It's supposed to be:

$\mathbf A \sim \mathbf H \implies \left\{{\mathbf x: \mathbf A \mathbf x = \mathbf 0}\right\} = \left\{{\mathbf x: \mathbf H \mathbf x = \mathbf 0}\right\}$

How embarrasing! --GFauxPas 03:31, 29 March 2012 (EDT)

Yep, nearly there ... all that needs to be done now is to indicate that performing the operations as you describe on the equations is "the same thing" as performing those operations on the matrices. You will also need to establish the meaning of "the same thing". --prime mover 04:29, 29 March 2012 (EDT)

You don't need to treat swapping to equations specially, since swapping rows $i$ and $j$ can be achieved e.g. by the following steps:

• $r_i \to -1 \cdot r_i$
• $r_i \to r_i +r_j$
• $r_j \to r_j -r_i$
• $r_i \to r_i +r_j$.

--Zahlenspieler (talk) 18:20, 17 January 2014 (UTC)