= Solution
{c}
A <system of linear equations> $T_\alpha x=b$ is solvable exactly when $b\in\operatorname{im}T_\alpha$. If $x_0$ is one solution, then
$$
T_\alpha x=b\iff T_\alpha(x-x_0)=0,
$$
so its solution set is $x_0+\ker T_\alpha$, a translate of the <kernel of a linear map>.
For $\alpha\ne2$, the <matrix> is an <invertible matrix>, and
$$
\boxed{\operatorname{im}T_\alpha=\mathbb R^3,\qquad\ker T_\alpha=\{0\}.}
$$
Every $b$ is in the <image of a linear map> and has exactly one preimage. For $\alpha=2$, its first two columns <span> the <image of a linear map>, while the corresponding <homogeneous linear system> gives the <kernel of a linear map>:
$$
\boxed{\operatorname{im}T_2=\{(u,v,w):u-v+w=0\},\qquad\ker T_2=\operatorname{span}\{(1,1,1)\}.}
$$
The prescribed $b=(-1,\beta,1)^T$ satisfies the displayed condition precisely when $\beta=0$. If $\beta\ne0$, it lies outside the <image of a linear map> and there is no solution. If $\beta=0$, it lies in the <image of a linear map>, and the nonzero <kernel of a linear map> supplies infinitely many solutions.
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