Find the values of $a$, $b$ and $c$ such that a matrix has infinite, unique, and no solutions.
$$x+y=0$$
$$y+z=0$$
$$x+z=0$$
$$ax+by+cz=0$$
We can't use determinants so I turned the equations into an augmented matrix:
\begin{matrix}
1&1&0&0 \\
0&1&1&0 \\
1&0&1&0 \\
a&b&c&0 \\
\end{matrix}
And then got this:
\begin{matrix}
1&1&0&0 \\
0&1&1&0 \\
0&a-b&-c&0 \\
0&0&2&0 \\
\end{matrix}
So in order to have a system with infinite solutions I said that $a-b=0$ and $-c=0$ meaning that $a=b$ and $c=0$. And to have a system with unique solution; $a-b=1$ so $a=1+b$. So far I don't know if that's correct, I don't really know how to solve these kinds of analytical problems and I have no idea how to proceed to find the values of $a.b.c$ so I have no solutions.
I would appreciate any help.
Best Answer
If we solve \begin{cases} x+y=0\\ y+z=0\\ z+x=0, \end{cases} then we get $x=y=z=0$. $(x,y,z)=(0,0,0)$ is also a solution of \begin{cases} x+y=0\\ y+z=0\\ z+x=0\\ ax+by+cz=0, \end{cases} but nonzero solution can't exist. Therefore, given system has unique solution for all $a,b,c$.
Meanwhile, we may simply say that given system of linear equations is equivalent to \begin{cases} x+y=0\\ y+z=0\\ z+x=0, \end{cases} because $ax+by+cz$ is linear combination of $x+y,y+z,z+x$. That can be justified since $$ ax+by+cz=\frac{a+b-c}{2}(x+y)+\frac{-a+b+c}{2}(y+z)+\frac{a-b+c}{2}(z+x). $$