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This is another question from Anton, Elementary Linear Algebra 9th Ed. The first system is:

$\begin{array}{1}x_{1}+3x_{2}-x_{3}=0\\ \,\,\qquad x_{2}-8x_{3}=0\\ \,\,\,\,\qquad\qquad 4x_{3}=0\end{array}$

Correction as demonstrated in the comments below. This system does not have a nontrivial solution.

The second system is:

$\begin{array}{1}a_{11}x_{1}+a_{12}x_{2}+a_{13}x_{3}=0\\ a_{21}x_{1}+a_{22}x_{2}+a_{23}x_{3}=0\end{array}$

For this second system it is an underdetermined homogeneous system of linear equations and so it has infinite solutions.

The third system is:

$\begin{array}{1}3x_{1}-2x_{2}=0\\ 6x_{1}-4x_{2}=0\end{array}$

This system is dependent linear equations, so they are collinear with infinite solutions.

The fourth system is:

$\begin{array}{1}2x_{1}-3x_{2}+4x_{3}-x_{4}=0\\ 7x_{1}+x_{2}-8x_{3}+9x_{4}=0\\ 2x_{1}+8x_{2}+x_{3}-x_{4}=0\end{array}$

With this fourth one, I cannot see any obvious sign to tell me there is more than the trivial solution. Is there something here that I am missing. All I can do at a glance is identify that it is not any of the other three cases. Maybe row-reduction might discover something else but this question is not supposed to use pencil and paper, i.e. no computation.

Thanks.

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  • $\begingroup$ Surely, in your first system, $x_3$ must be zero? $\endgroup$
    – Angina Seng
    Jun 11, 2017 at 6:27
  • $\begingroup$ oops, yes of course, sorry about that. Let me look at that one more closely. $\endgroup$
    – Bucephalus
    Jun 11, 2017 at 6:27
  • $\begingroup$ Ok thanks @LordSharktheUnknown. I can see that this one has only the trivial solution. My bad. Will correct the question. $\endgroup$
    – Bucephalus
    Jun 11, 2017 at 6:29
  • $\begingroup$ The fourth and second systems are of the same type. $\endgroup$
    – amd
    Jun 11, 2017 at 7:19
  • $\begingroup$ Thanks for that @amd. $\endgroup$
    – Bucephalus
    Jun 11, 2017 at 7:23

1 Answer 1

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Hint:-

If you are given m homogeneous equations in n unknowns and $m < n$, then you always get infinitely many solutions, since at least one variable becomes free.

For a homogeneous system, if you have m equations in n unknowns with $m = n$, then you get only the trivial solution.

I hope this helps you solve the problems.

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  • $\begingroup$ Ahh, I did not now this maxim @AniruddhaDeshmukh. Thankyou. $\endgroup$
    – Bucephalus
    Jun 11, 2017 at 6:34
  • $\begingroup$ Actually I did know that didn't I. I used that axiom in the second case. I didn't recognise to use it in the fourth case. I need a coffee. $\endgroup$
    – Bucephalus
    Jun 11, 2017 at 6:37

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