(i) no solution (ii) a unique solution (iii) infinitely many solutions with one parameter (iv) infinitely many solutions with 2 parameters
$$ \begin{matrix} c^2-c & 0 & c & c^2 \\ 0 & 1-c^2 & c^2 & -c\\ 0 & c^2-1 & c & c\\ c^2-c & 0 & c & 2c\\ \end{matrix} $$
I understand that we are supposed to use Gaussian elimination and solve this linear system by considering entries. Is it possible for this linear system to have infinitely many solutions?
$R_1\times-1+R_4 \longleftrightarrow R_4$ $$ \begin{pmatrix} c^2-c & 0 & c & c^2 \\ 0 & 1-c^2 & c^2 & -c\\ 0 & c^2-1 & c & c\\ 0 & 0 & 0 & c(2-c)\\ \end{pmatrix} $$ We see if $c \neq0,2$ then the original system has no system. However, if $c=0$, we arrive at the system$$ \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{pmatrix} $$ which has infinitely many solutions with two parameters: $$ \begin{pmatrix} x_1\\ x_2\\ x_3\\ \end{pmatrix} =x_1\begin{pmatrix} 1\\ 0\\ 0\\ \end{pmatrix}+x_3\begin{pmatrix} 0\\ 0\\ 1\\ \end{pmatrix} $$ On the other hand, if $c=2$ we arrive at $$ \begin{pmatrix} 2 & 0 & 2 & 4 \\ 0 & -3 & 4 & -2\\ 0 & 3 & 2 & 2\\ 0 & 0 & 0 & 0\\ \end{pmatrix} $$ $R_2+R_3 \longleftrightarrow R_3$ $$ \begin{pmatrix} 2 & 0 & 2 & 4 \\ 0 & -3 & 4 & -2\\ 0 & 0 & 6 & 0\\ 0 & 0 & 0 & 0\\ \end{pmatrix} $$ which possesses a unique solution
$$ \begin{pmatrix} x_1\\ x_2\\ x_3\\ \end{pmatrix} =\begin{pmatrix} 2\\ 2/3\\ 0\\ \end{pmatrix}$$ There is no value of $c$ for which this system will have infinitely many solution with one parameter.