0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Surely, in your first system, $x_3$ must be zero? $\endgroup$ – Lord Shark the Unknown Jun 11 '17 at 6:27
  • $\begingroup$ oops, yes of course, sorry about that. Let me look at that one more closely. $\endgroup$ – Bucephalus Jun 11 '17 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 '17 at 6:29
  • $\begingroup$ The fourth and second systems are of the same type. $\endgroup$ – amd Jun 11 '17 at 7:19
  • $\begingroup$ Thanks for that @amd. $\endgroup$ – Bucephalus Jun 11 '17 at 7:23
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Ahh, I did not now this maxim @AniruddhaDeshmukh. Thankyou. $\endgroup$ – Bucephalus Jun 11 '17 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 '17 at 6:37

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.