0
$\begingroup$

Consider the following system of linear equation:

\begin{align} 2a + 4b &= a + 3c\\ 2a + 3b &= 4a + 2b\\ 4a + 2b &= b + nc \end{align}

for $a,b,c \in \mathbf{R}_{+}$.

How do I find the value of $n \in \mathbf{N}$, assuming there is a unique solution for $(a,b,c)$?

(I guess substitution is the way to go but I can't figure it out.)

$\endgroup$
  • $\begingroup$ You're not assuming it has a unique solution. You're assuming it has infinitely many solutions i.e. you're assuming it has a solution other than (0,0,0). $\endgroup$ – peter.petrov Mar 6 '16 at 17:50
2
$\begingroup$

From the 2nd equation, $b=2a$.

From the 1st equation, $9a=3c$ so $c=3a$.

From the 3rd equation, $8a=2a+3na$ so $3n=6$ (since $a\ne0$) and $n=2$.

$\endgroup$
  • $\begingroup$ Nice and simple! $\endgroup$ – Svend Tveskæg Mar 6 '16 at 17:47
1
$\begingroup$

The determinant {{1,4,-3},{-2, 1, 0},{4, 1, -n}} is $18-9n$.
What you want is this determinant to be $0$. So $n=2$.

determinant

$\endgroup$
0
$\begingroup$

See as $0$ is the solution any value of $n $ satisfies it but now if we have some solution apart from $0$ then use determinants. For this system the determinant of third order should be equal to $0$ if you plug values you get a solution ie $9n=18$ thus $n=2$ has solution which is not $[0,0,0]$

$\endgroup$
  • $\begingroup$ Can you write it out, please? I don't really get it ... sorry. $\endgroup$ – Svend Tveskæg Mar 6 '16 at 17:45
  • $\begingroup$ I am not sure where $9n=18$ came from. $\endgroup$ – peter.petrov Mar 6 '16 at 17:45
  • $\begingroup$ Yes its $9n-18$ for your last row it should be $4,2,n$ then see $\endgroup$ – Archis Welankar Mar 6 '16 at 17:49
  • $\begingroup$ No, it's (4,1,-n) my last row. OK, got it, thanks. $\endgroup$ – peter.petrov Mar 6 '16 at 17:51

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.