Homogeneous system with solutions for all a,b,c

Question

Prove that for all $a,b,c \in \mathbb R$ the next system has non-zero solutions: $$x+ay+(b+c)z=0$$

$$x+by+(c+a)z=0 $$

$$x+cy+(a+b)z=0 $$

Choosing $\delta_p=b-a$ it won't be for all $a,b,c \in \mathbb R$.

Answer 1

The solutions are given by $$ x= - z(a + b + c),\; y=z $$ for arbitrary $z$.

Answer 2

Hint

We can rewrite these equation as$$x+a(y-z)=-(a+b+c)z\\x+b(y-z)=-(a+b+c)z\\x+c(y-z)=-(a+b+c)z$$from which we obtain$$a(y-z)=b(y-z)=c(y-z)$$Now what happens for cases where $a=b=c$ and $!(a=b=c)$?

Answer 3

Consider the system as $$\begin{bmatrix}1&a&b+c\\1&b&a+c\\1&c&a+b\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}.$$ Using row reduction, we get $$\begin{bmatrix}1&a&b+c\\1&b&a+c\\1&c&a+b\end{bmatrix} \longrightarrow \begin{bmatrix}1&a&b+c\\0&b-a&a-b\\0&c-a&a-c\end{bmatrix}.$$ If either $a=b$ or $a=c$ or both, then this system will have non-trivial solutions (as one of the rows will become a zero row).

If $a \neq b$ and $c \neq a$, then the last rows are still going to be the same, hence once again we can get a zero row after one more row operation. Thus the system will still have non-trivial solution.