= Solution
Use the <eigenvector> <basis> $v_1=(-1,0,1)^T$, $v_2=(1,1,0)^T$, $v_3=(1,1,1)^T$. The forcing has decomposition $b=-v_1+3v_3$. If $x=c_1v_1+c_2v_2+c_3v_3$, the <linear system> reduces to
$$
(1-\lambda)c_1=-1,\qquad(2-\lambda)c_2=0,\qquad(3-\lambda)c_3=3.
$$
For $\lambda=0$, all coefficients are determined, giving
$$
\boxed{x=(2,1,0)^T.}
$$
For $\lambda=1$, the first equation would say $0=-1$, so \b[there are no solutions]. For $\lambda=2$, $c_1=1$, $c_3=3$, and $c_2$ is arbitrary. Thus \b[all solutions] are
$$
\boxed{x=(2,3,4)^T+t(1,1,0)^T,\qquad t\in\mathbb R.}
$$
These outcomes illustrate respectively an invertible <matrix>, incompatible forcing outside its <image of a linear map>, and a compatible forcing with a nontrivial <kernel of a linear map>.
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