# adding rows together linear algebra

In the context of constant coefficient linear matrices, why is it okay to add one row to another? I understand multiplying an equation by a constant, and even adding a number to both sides of an equation. But, how does adding one equation to another not change it? I have used this plenty of times in my study, but it only just occurred to me that this isn't the same as, say,

$q = 2 \cdot x$

and

$q + 5 = 2 \cdot x + 5$

So how does it work, that adding/subtracting two equations together doesn't change the solution of

$Ax = b$

• It's basically a consequence of $a=b\ \&\ c=d\implies a+c=b+d$. – Akiva Weinberger Feb 1 '16 at 0:27
• @AkivaWeinberger: no, it's a consequence of $a = b \mathrel{\&} c = d \iff a = b \mathrel{\&} a + c = b + d$ – Rob Arthan Feb 1 '16 at 0:32
• @RobArthan Yes, that implies it more directly. :) – Akiva Weinberger Feb 1 '16 at 0:33
• @AkivaWeinberger: cheers! it's a game of two halves $\ddot{\smile}$! – Rob Arthan Feb 1 '16 at 0:35
• Thank you for asking the question: it's really good that you aren't just swallowing mathematics without thinking about it! A lot of my students wouldn't be able to tell me why adding equations is OK! – Frentos Feb 5 '16 at 12:42

Suppose we have two equations $\left\{\begin{array}{rrrcl}2x&+3y&-z&=&4\\x&-2y&+2z&=&6\end{array}\right.$

We can add $6$ to both sides of the top equation in the usual way like this: $$2x + 3y - z + 6 = 4 + 6$$ but because of the 2nd equation, $6$ is the same as $x-2y+2z$, so we can write: $$2x + 3y - z + (x-2y+2z) = 4 + 6$$

That (hopefully!) convinces us that adding equations is OK.

But why doesn't this change the solution space? It may help to think about the true/false value of the equations for different $(x,y,z)$.

• A solution makes both equations true; a non-solution makes at least one of the equations false.

• If we consider a solution, both equations are true for those values; if both equations are true, the sum of the equations is true (read over the example above thinking about truth/falsehood), so no solutions are lost.

If we consider a non-solution, the situation is a bit trickier. We might start with $$\begin{array}{rcl}y&=&2x\\y&=&4x\\\hline 2y&=&6x\end{array}$$ The non-solution $x=1,y=3$ makes both the first two equations false, but makes the last equation true! How do we know non-solutions can't sneak in this way? I can't think of a succinct explanation of why this is not a problem in general, but:

• If you shift your point of view to logic, it's not a problem that false premises can lead to any conclusion, including true conclusions)
• In the context of matrices or systems of equations it's not a problem because you'll always have some equation hanging around that the non-solution makes false. In a way, you can view reducing an augmented matrix to row echelon form and getting an inconsistency as pushing the falsehood of any candidate solution into one equation like $0x+0y+0z=5$. Consider a reduced matrix with one solution: $\left|\begin{array}{ccc|c}1&0&0&2\\0&1&0&3\\0&0&1&-4\end{array}\right|$. The last equation in the matrix, $z=-4$, is true for a non-solution like $x=0,y=1,z=-4$, but it's ruled out when you consider the other equations.
• This is perfect. Thank you for a very informative answer. – justthom8 Feb 5 '16 at 19:47