Solution
= Solution
The coefficient <matrix> has <determinant> $-1$, so it is an <invertible matrix>. Its <homogeneous linear system> therefore has only the zero solution:
$$
\boxed{x=y=z=0}.
$$
One can also subtract the first equation from the second to get $z=0$, then subtract the first from the third to get $y+2z=0$, giving $y=0$ and finally $x=0$. Thus \b[there are no nonzero solutions].