I am asked to the find the value $c$ such that the system has a solution other than $(0, 0, 0)$.
This is the linear system:
\begin{array}{rcrcrcl} cx & & & + & 4z& = & 0\\ & &7y & - &70z& = & 0\\ 10x& - & y & & & = & 0 \end{array}
The answer I got was $c=-4$ but I'm not sure if that's the correct answer.
I tried applying row-reduced echelon form to the system but it kept bringing me in loops, as there wasn't enough variables to eliminate. Does anyone know how to find the value for $c$?
$\endgroup$ 24 Answers
$\begingroup$The trick to this problem is keeping track of the $c$ and making sure you are not treating it as a variable to solve for like $x$, $y$, or $z$.
First, let's number the equations: $$\tag1 cx+4z=0$$ $$\tag2 7y-70z=0$$ $$\tag3 10x-y=0$$
We can immediately see that it might be helpful for us to divide by $7$ in the second equation: $$\tag2 y-10z=0$$
Now, let's try applying row reduction, as you did. However, instead of using a matrix, I think in this scenario, since a lot of the variables have a coefficient of $0$ in the equations, it will be easier simply to try to get rid of variables (which is basically the same as row reduction; a good example of this is on Wikipedia) rather than using a matrix and having to deal with so many $0$s.
First, we need to get rid of $x$ from all equations except the first. The only equation with $x$ besides the first one is the last one, $10x-y=0$.We know that $cx+4z=0$, so we can multiply this by $\frac{10}{c}$ to get $10x+\frac{40}{c}z=0$. If we subtract equation $(3)$ by this new equation, we get: $$\tag3 -y-\frac{40}{c}z=0$$ It will probably be helpful for us to multiply everything by $-1$ to get rid of the $-$s: $$\tag3 y+\frac{40}{c}z=0$$
Now, we want to get rid of $y$ from all of the equations except the second. The only equation with $y$ except the second is the third, which is stated above. The second equation is $y-10z=0$, so if we subtract the third equation by the second equation, we get the following: $$\tag3 \left(\frac{40}{c}+10\right)z=0$$
Now, clearly, either $z=0$ or $\frac{40}{c}+10=0$. If $z=0$, then it's easy to find that $x=0$ and $y=0$ by substitution, so our solution is $(0, 0, 0)$. That's not what we wanted! However, if $\frac{40}{c}+10=0$, then $c=-4$, but $z$ could be anything. If we continue with row reduction and get rid of $z$ from the first and second equation, then we still find that $x=0$ and $y=0$, but now we find that $(0, 0, z)$ is a solution for all $z \in \Bbb{R}$. Thus, this is why $c=-4$ is the only way there can be a solution other than $(0, 0, 0)$
$\endgroup$ 2 $\begingroup$$$ \begin{array} \\ y=10x \\ 70x-70z=0 \implies x=z \\ cx+4x=0 \implies c=-4 \text{ if x is not zero} \end{array}$$ so if $c=-4$ the non-trivial solutions are constant multiples of $(1,10,1)$
$\endgroup$ $\begingroup$Combining the last two equations gives $70(x-z)=0$. We want to make this consistent with $cx+4z=0$ for $x,z\neq 0$. If $c=-4$ then every value of $x$ has a corresponding $z$ that satisfies the equation. Thus your answer works.
$\endgroup$ $\begingroup$In doing your elimination, you should treat $c$ as a constant and eliminate $x,y,z$ The second gives $y=10z$, which gives $x=z$ from the third. Then the first becomes $(c+4)z=0$ This will force $z=0$ unless $c=-4$ as you surmise. If $c=-4$, then $z$ can be anything and you can find values for $x,y$ in terms of $z$
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