If $n$ is a real number, then the simultaneous system 
$nx+y = 1$
$ny+z = 1$
$x+nz = 1$
has no solution if and only if $n$ is equal to 
$\textbf{(A)}\ -1\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 0\text{ or }1\qquad\textbf{(E)}\ \frac{1}2$
Solution 1
Add up all three equations to get
\[(n+1)(x+y+z) = 3\]
If $n = -1$, then equation results in $0 = 3$.  That has no solutions, so the answer is $\boxed{\textbf{(A)}}$.

Solution 2
From the first equation,
\[y = 1 - nx\]
Substitute that in the second equation to get
\[n(1-nx) + z = 1\]
\[z = 1 - n + n^2 x\]
Substitute that in the third equation to get
\[x + n(1-n+n^2 x) = 1\]
\[(1+n^3)x + n - n^2 = 1\]
If $n = -1$, then equation results in $-2 = 1$.  That has no solutions, so the answer is $\boxed{\textbf{(A)}}$.