Prove that system of equations has no solution when condition is met Can I have some help with the following question?
Show that the system of equations $$\begin{bmatrix}1&2&-1\\2&-3&2\\-1&12&-7\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$$
has no solutions unless parameters $a, b, c$ are such that $3a - 2b - c = 0$. What is the general solution if $a = b = c = 1$?
I know that a system of equations has the determinant of A equal 0, which in my case is of the first matrix in the equation, than the system does not have a solution, which is the case here. If I do the characteristic determinants (Dx, Dy, and Dz), I do indeed get $3a - 2b - c$, but still do not understand how to link that to the task of the problem. Any advice, or tips for what I am missing is greately appreciated.
 A: I think I figured it out. If one row of matrix A is all 0s, than it means that a variable can take any value. Furthermore, if the coresponding number from the vector is 0, than the system has one redundant equation which can be taken away, and than solve the remaining two equations by taking one of the variables to the right side as a parameter. If the coresponding number from the vector is non-zero (say n), than the system has no solution, because what we obtain is $0x+0y+0z=n$, but n is not 0, so it can't have a solution. But this is just more like an observation, I'd appreciate if someone would properly explain/develop on what I researched/figured out.
A: Since you have managed to show that $3a - 2b - c = 0$ you have therefore answered the first part of your question and there is no need to "understand how to link that to the task of the problem". 
So I will answer the second part of your question; That is, to find the general solution when $a=b=c=1$:
  $$\begin{bmatrix}1&2&-1\\2&-3&2\\-1&12&-7\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix}\implies\left\{ 
\begin{array}{ll}
x+2y-z & =1 & \text{(1)} \\ 
2x-3y+2z&=1 & \text{(2)} \\ 
-x+12y-7z&=1 & \text{(3)}
\end{array}
\right.$$
To solve this system of equations (find the values of $x$,$y$ and $z$ that satisfy the system of equations) I will use Row Reduction but I won't write it out as an Augmented Matrix as the steps are harder to explain: The objective of Row Reduction is to use the first row $(1)$ to clear the rest of the first column (so that only zero's remain for the other $x$ coefficients); next use the new second row $(4)$ to clear the rest of the second column (so that only zero's remain for the other $y$ coefficients); etc. 
Multiply $(1)$ by $-2$ and add to $(2)$. Then add $(1)$ to $(3)$. 
$$
\implies\left\{
\begin{array}{rc}
x+2y-z & = 1 & \text{(1)}\\ 
   -7y + 4z & = -1 &\text{(4)} \\ 
14y-8z &=2 &\text{(5)}
\end{array} 
\right.
$$
Multiply $(4)$ by $2$ and add to $(5)$. Then multiply $(4)$ by $\dfrac{2}{7}$ and add to $(1)$.
$$
\implies\left\{
\begin{array}{ll}
x+\frac17 z & = \frac57 & \text{(6)}\\ 
   -7y + 4z & = -1 &\text{(4)} \\ 
 &=0 &\text{(7)}
\end{array} 
\right.
$$
At this point the bottom row is completely clear. So the rows of the matrix have been reduced as far as possible. To tidy up a bit we divide $(4)$ by $-7$.
$$
\implies\left\{
\begin{array}{ll}
x+\frac17 z & = \frac57 & \text{(6)}\\ 
   y  -\frac47 z & = \frac17 &\text{(8)} 
\end{array} 
\right.
$$
Now let $z=\lambda$ and the general parametric solution (in terms of $\lambda$) is $$\color{blue}{x=\frac17 \left(5-\lambda \right)}$$ and $$\color{#F80}{y=\frac17 \left(1+4\lambda \right)}$$
