# System of Linear Equations with integer Coefficients

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.)

• 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). – peter.petrov Mar 6 '16 at 17:50

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 $n=2$.

• Nice and simple! – Svend Tveskæg Mar 6 '16 at 17:47

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

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 $[0,0,0]$

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