Gaussian Elimination and Matrix question 
Given the system of equations:
$$x_1        +   5x_3  = 1$$ 
$$2x_1  –   x_2 + 7x_3  = 7 $$
$$2x_2 + cx_3  = – 1 $$
where $c$ is a constant.
(the number behind the $x$ is supposed to be a number lower than the $x$,
  I just don't know how to use the system in here to write equations
  yet). 
(a) Write down the extended matrix. Find the solution using Gaussian
  elimination method. For what values ​​of $c$ and equation unique
  solution? Interpret the meaning of other values for $c$ (e.g. forbidden
  values).
(b) Assume that the system of equations above can be written in matrix
  form $Ax = B$. Let $c = 3$ Set up $A, x, B$. Calculate the determinant of
  $A$.
(c) Compute the inverse of $A$ by means of the adjoint matrix (matrix of
  cofactors). Use the result for $A -1$ to find the solution to equation
  set. 

I'm not very good at math but I've tried this one. My books is really bad at explaining things and I don't really know anyone who is good at math. I would very much appreciate help with this problem.
I've read up on homework questions in here and also sent my professor an email, he said it was fine as long as I afterwards were able to show that I understood it, not just copy pasted of someone else. 
(b) and (c) I'm really in the dark about these two, but I think I've understood somewhat how to do (a).
The extended matrix:
$$\begin{bmatrix}
1  & 0 & 5 &  1 \\  
2  & -1 & 7  & 7 \\ 
0  & 2 &  c & -1  \\
\end{bmatrix}
$$
And in (a) I multiplied the first line and mixed it with the second to remove ($2x_1$), and in the third line (where my problem is) I got $(c-6)x_3 = 9$. 
So, I'd love some help with $a-c$, but if that's not allowed It would be very helpful if someone could explain what to do with the $c$ in the third line. What I understood is that if I manage to get $x_3 = "something"$ I could use that in the other lines to find $x_1$ and $x_2$ and that would mean that I've solved it using Gaussian elimination right? 

Your system of equation give the following matrix as you did: $$\begin{bmatrix} 1  & 0 & 5 &  |1 \\  
2  & -1 & 7  & |7 \\ 
0  & 2 &  c & | -1  \\
\end{bmatrix}$$ Now by doing $-2R_1+R_2\rightarrow R_2$ you have $$\begin{bmatrix}
1  & 0 & 5 &  |1 \\  
0  & -1 & -3 &| 5 \\ 
0  & 2 &  c & |-1  \\
\end{bmatrix}$$ and by $2R_2+R_3\rightarrow R_3$ you have $$\begin{bmatrix}
1  & 0 & 5 &  |1 \\  
0  & -1 & -3 &| 5 \\ 
0  & 0 &  c-6 & |9  \\
\end{bmatrix}$$ Now you can arrange a new system of equations by using the final matrix exactly as you get the first extended matrix. I mean: $$x_1+5x_3=1\\-x_2-3x_3=5\\(c-6)x_3=9$$ Suppose that $c=6$ then what happens? A contradiction! In this case the system has no solution as form $(x_1,x_2,x_3)$. On the other hand you can read the determinant from the final matrix. It is $1*(-1)*(c-6)=6-c$. If you want your non homogenous system has a unique solution; the determinant should not be zero. It means that $c$ must be any number than $6$.
