= Solution
{c}
In <morphoelasticity>, the <deformation gradient> factors as $\mathbf F=\mathbf A\mathbf G$, where $\mathbf G$ accounts for local stress-free growth and $\mathbf A$ accounts for the subsequent elastic accommodation. <Elastic incompressibility> means that the latter preserves volume:
$$
\det\mathbf A=1.
$$
It does not forbid volume increase by growth: here $\det\mathbf F=\det\mathbf G=\gamma$.
For the homogeneous diagonal deformation, the two unloaded transverse directions are equivalent. Taking the transversely symmetric branch implicit in the paper's hint gives $\alpha_1=\alpha_2>0$. Set $\alpha_3=\alpha>0$. The <determinant> constraint is then
$$
\alpha_1^2\alpha=1,
$$
so
$$
\boxed{\mathbf A=\operatorname{diag}\left(\alpha^{-1/2},\alpha^{-1/2},\alpha\right).}
$$
Equality of the transverse stretches uses that symmetry assumption; <elastic incompressibility> alone would only give $\alpha_1\alpha_2\alpha_3=1$. The total <deformation gradient> and the actual axial length are
$$
\mathbf F=\operatorname{diag}\left(\alpha^{-1/2},\alpha^{-1/2},\alpha\gamma\right),
\qquad \ell=\alpha\gamma,
$$
because the initial <cuboid> is a unit cube.
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