# Find the values of a,b,c such that a matrix has infinite, unique, and no solutions.

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.

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Is that second line supposed to be $y+x$ or should it be $y+z$? –  Mike Feb 27 '13 at 1:57
First, terminology: matrices don't have solutions. A system of equations may, or may not have solutions. Second, what you have is a homogeneous system, so "no solution" is impossible --- can you see why it's guaranteed that there is at least one solution? –  Gerry Myerson Feb 27 '13 at 1:58
@Mike Sorry, my bad. –  ChairOTP Feb 27 '13 at 1:59
@Gerry Myerson, again my bad. I can see now why. But the others are correct? –  ChairOTP Feb 27 '13 at 2:00
If you ignore the 4th equation, how many solutions does the system have? Then, can remembering the 4th equation (thus imposing an additional condition on solutions) increase the number of solutions? –  Gerry Myerson Feb 27 '13 at 2:32