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.)
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$\begingroup$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 $n=2$.
$\endgroup$ 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 $n=2$.
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 $[0,0,0]$
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