I'm just starting out, so please bear with me.

While solving systems of equations in linear algebra, can you straight-up add, if two factors cancel?

My problem:

$$x_1 + 3x_2 -x_3 +x_4 +2x_5 = 2,$$

$$2x_1 -x_2 +x_3 -x_4 + x_5 = -2,$$

$$4x_1 +2x_2 +2x_3 -x_4 -x_5 = 0.$$

I thought that I could multiply 1 by equation 2, and then add that to equation 1, which would cancel both the $x_3$ and the $x_4$ factors for that line. Is that permissible? My book says to begin with the $x_1$ factors, so I'm not sure. Thanks.

  • $\begingroup$ Of course you can, you get $3x_1 + 2x_2 + 3x_5= 0$ that is a new true equation! $\endgroup$
    – Maffred
    Jan 29, 2016 at 3:11
  • $\begingroup$ By the way it's more clear if you make a scale reduction starting from $x_1$, ie keep the first line, substitute the second line with "second line minus two times the first". And substitute the 3rd one with "3rd minus 4 times the first". You get a new system. Repeat and rinse with $x_2$. $\endgroup$
    – Maffred
    Jan 29, 2016 at 3:16
  • 1
    $\begingroup$ Your book's way of working is more systematic and will work in all cases. Your idea works when you see an operation which will greatly simplify your system. But I've seen too many students' calculations which would take two pages and get almost nowhere. After a few years teaching, I insisted on the systematic approach. Also this approach can be easily formalised in a computer program. $\endgroup$
    – MasB
    Jan 29, 2016 at 4:20
  • $\begingroup$ I'm still confused. One factor becomes 0 and then in the next step it changes to non-zero. So, I don't know. I hope to catch my professor tomorrow. $\endgroup$ Jan 29, 2016 at 5:40

1 Answer 1


What you suggest is just fine. But doing what your book suggests is not a bad idea either. Since you have three equations and five unknowns, you will only be able to solve for three of the variables (at most) in terms of the other two anyway.


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