I am doing some dimensional analysis from an Applied Mathematics textbook.

They present a simple 4 x 6 matrix, which gives rise to the following equations:

$a_1 - a_5 = 0$

$a_2 + 2a_5 - 3a_6 = 0$

$a_3 - a_6 = 0$

$a_4 + a_6 = 0$

The text then states that, through 'conventional linear algebra methods', the following solutions are obtained:

$a_1 = -\frac{1}{2}, a_2 = 1, a_3 = a_4 = 0, a_5 = -\frac{1}{2}, a_6 = 0$


$a_1 = \frac{3}{2}, a_2 = 0, a_3 = 1, a_4 = -1, a_5 = \frac{3}{2}, a_6 = 1$

I don't see the logic in moving from the homogeneous system to the given sets of linearly independent solutions. Is this a trick I haven't been taught? Is it a matter of assigning arbitrary values to the free variables $a_5$ and $a_6$?

Any help is enormously appreciated.


  • $\begingroup$ So do you mean that in the first set of solutions, that they have deliberately assigned $a_5 = -\frac{1}{2}$? How did they come up with that number? They talk about those two sets of solutions as though they're the \textit{only} correct ones. Argh $\endgroup$ – Victoria Mar 7 '15 at 1:39
  • $\begingroup$ For instance, assignign $a_5 = 0$ and $a_6 = 0$ makes everything into a zero. But that can't be right surely? $\endgroup$ – Victoria Mar 7 '15 at 1:42
  • $\begingroup$ Ah okay, thanks for helping me out $\endgroup$ – Victoria Mar 7 '15 at 1:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.