$[A \vdots B]$ =$\begin{bmatrix} 1 & -1 & -1& \vdots &0 \\ 1& 1& 1 & \vdots& 46\\ 1& -2& 1&\vdots &16 \\ -1& -1& 2& \vdots&-7\end{bmatrix}$
R2->R2-R1 , R3->R3-R1 and R4->R1+R4
=$\begin{bmatrix} 1 & -1 & -1&\vdots & 0\\ 0& 2& 2 & \vdots&46 \\ 0& -1& 2&\vdots & 16\\ 0& -2& 1& \vdots & -7 \end{bmatrix}$ R2->R2/2
=$\begin{bmatrix} 1 & -1 & -1& \vdots & 0 \\ 0& 1& 1 & \vdots& 23\\ 0& -1& 2&\vdots &16 \\ 0& -2& 1& \vdots & -7 \end{bmatrix}$R3->R2+R3 and R4->2R2+R4
=$\begin{bmatrix} 1 & -1 & -1& \vdots &0 \\ 0& 1& 1 & \vdots & 23\\ 0& 0& 3& \vdots & 39\\ 0& 0& 3& \vdots & 39 \end{bmatrix}$ R4->R4-R3
=$\begin{bmatrix} 1 & -1 & -1& \vdots &0 \\ 0& 1& 1 & \vdots & 23\\ 0& 0& 3& \vdots & 39\\ 0& 0& 0& \vdots & 0 \end{bmatrix}$
Here
Rank[A]=Rank$[A \vdots B]$ = n(Number of variables)
Hence,Unique solution is possible.
Hence,Option(C)Exactly one solution is the correct choice.