= Solution
{c}
For $\alpha\ne2$, $A'=I_3$, whose <row space> is $\mathbb R^3=\operatorname{im}T_\alpha$. For $\alpha=2$, <elementary row operations> give the <reduced row echelon form>
$$
A'=\begin{pmatrix}1&0&-1\\0&1&-1\\0&0&0\end{pmatrix}.
$$
Its <row space> is $\{(u,v,w):u+v+w=0\}$, whereas the <image of a linear map> is $\{(u,v,w):u-v+w=0\}$. These differ: $(0,1,-1)$ belongs to the former but not the latter. Thus \b[(A) is true exactly when $\alpha\ne2$]. Rows are being identified with <vectors> in $\mathbb R^3$ in this comparison.
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