$$\begin{cases} 2x_1-x_2= dx_1 \\ 2x_1-x_2+x_3=dx_2 \\ -2x_1+2x_2+x_3=dx_3 \end{cases} $$
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)?
Using the Gaussian elimination method to solve a set of linear equations,
From the equations, you have given,
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}