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}
If $a^2 - 4 \neq 0$, then yes, there is a unique solution. Dividing by the non-zero number $a^2 - 4$ (whatever it is equal to) will put a pivot in each column, and finding the solution from there involves back-substitution.
If $a^2 - 4 = 0$, then we don't have a pivot in each (non-augmented) column. This doesn't mean there is no solution necessarily; there might be infinitely many. To have any solutions at all, we must verify that, once in row-echelon form, every $0$ row extends to the augmented column. That is, we have no rows of the form
$$[\begin{array}{cccc|c} 0 & 0 & \cdots & 0 & 2\end{array}]$$
for example. Anything non-zero in the augmented column will mean there's no solutions.
In this case, when $a^2 - 4 = 0$, then we get $a^2 + 2 = 6 \neq 0$ (in fact, $a^2 + 2$ is never equal to $0$ for $a \in \Bbb{R}$). So, indeed, there will be no solutions when $a^2 - 4 = 0$. Please note that this wasn't guaranteed by the zero row; we needed to check the augmented column was non-zero.
Infinitely many solutions occur when the system is consistent (i.e. at least one solution exists), and there is a column without a pivot. When considering a square matrix of coefficients, this must be accompanied by a zero row (extending into the augmented column). This can never happen in the given matrix, as we have just shown.
Hope that helps!
Best Answer
These matrices are augmented, which means the last column should be the numbers on the right side of the equation.
(a) Is consistent but not infinitely many solutions. Its equation would be $\blacksquare x_1=*, 0x_1=0$. So there should be only one solution.
(b) Is obviously inconsistent because $\blacksquare$ cannot be zero, but $0x_1$ will have to be zero.
(c) There should be only one unique solution, the first and second rows are obviously not the same with each other, so two useful equations for two unknown variables.
(d) inconsistent, similar to (b)
(e) You are right it is infinitely many solutions, because only two useful equations to solve for three unknown variables.