= Solution
Write a vector in the domain as $(u,v,w,t)^T$. The first and third equations defining the <kernel of a linear map> imply $v=2\alpha u-2t$. Substitution into the second equation then gives
$$
2(\alpha-1)\bigl(t-(\alpha+1)u\bigr)=0.
$$
For $\alpha\ne1$, this forces $t=(\alpha+1)u$, $v=-2u$ and $w=3u$. Hence
$$
\boxed{\ker M_\alpha=\operatorname{span}\{(1,-2,3,\alpha+1)^T\},\qquad\operatorname{im}M_\alpha=\mathbb R^3\quad(\alpha\ne1).}
$$
The <kernel of a linear map> has dimension one, so the <rank-nullity theorem> gives rank three and proves the assertion about the <image of a linear map>. Alternatively, the minor formed from columns $2,3,4$ has <determinant> $2(\alpha-1)$, including a nonzero value at $\alpha=-1$.
For $\alpha=1$, the parameters $u,t$ are free and $v=2u-2t$, $w=-3u+3t$. Thus
$$
\boxed{\ker M_1=\operatorname{span}\{(1,2,-3,0)^T,(0,-2,3,1)^T\}.}
$$
The third row is the first row minus the second. The <image of a linear map> is therefore contained in $y_1-y_2-y_3=0$. The first and third columns, $(1,2,-1)^T$ and $(1,0,1)^T$, are linearly independent, so the <matrix rank> is two and this containment is equality:
$$
\boxed{\operatorname{im}M_1=\{y\in\mathbb R^3:y_1-y_2-y_3=0\}=\operatorname{span}\{(1,2,-1)^T,(1,0,1)^T\}.}
$$
This is the sole <kernel jump in a parameter-dependent linear map>; there is no further exceptional case at $\alpha=-1$.
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