How do I solve 6b) and 6c) if my solution for 6a) is a consistent system of linear equations? $$\begin{cases}
2x_1-x_2= dx_1 \\ 
2x_1-x_2+x_3=dx_2 \\ 
-2x_1+2x_2+x_3=dx_3
\end{cases}
$$
a) Is it possible for the system to be inconsistent? Explain?
b) For what values of d will the system have infinitely many solutions?
c) Solve the system when it has infinitely many solutions?
For my solution in part a), 
$$ \left[
\begin{array}{ccc|c}
  1&0&0&0\\
  0&1&0&0\\
  0&0&1&0
\end{array}
\right] $$
Hence, it is a unique set of solutions i.e. the system cannot be inconsistent. So how is it possible to get infinitely many solutions in 6b) and 6c)?
 A: Using the Gaussian elimination method to solve a set of linear equations,
From the equations, you have given, 

\begin{cases}
2x_1-x_2= dx_1 \\ 
2x_1-x_2+x_3=dx_2 \\ 
-2x_1+2x_2+x_3=dx_3
\end{cases}

We can arrive at this augmented matrix,
\begin{bmatrix}
\begin{array}{ccc|c}
  -d+2&-1&0&0\\
  2&-1-d&1&0\\
  -2&2&1-d&0
\end{array}
\end{bmatrix}
Using row transformations,
\begin{bmatrix}
\begin{array}{ccc|c}
  -2   &  2   & 1-d &0 \\
   2   & -1-d & 1   &0\\
  -d+2 & -1   & 0   &0\\
\end{array}
\end{bmatrix}
\begin{equation}
\downarrow
\end{equation}
\begin{bmatrix}
\begin{array}{ccc|c}
  -2   &  2   & 1-d &0 \\
   0   &  1-d & 2-d   &0\\
   0   &  0   & (1-d)(2-d) -2(2-d)   &0\\
\end{array}
\end{bmatrix}
For it to have infinite solutions, 
\begin{equation}
(1-d)(2-d) -2(2-d) = 0
\end{equation}
\begin{equation}
d = 2, -1
\end{equation}
If $d$ takes the above value, then you will end up with a free variable ($x_3$)
\begin{equation}
x_2 = \frac{(d-2)x_3}{1-d}
\end{equation} 
\begin{equation}
x_1 = \frac{1}{2}(2x_2 + (1-d)x_3)
\end{equation}
