\\
0 & 0 &\gamma
 \end{pmatrix} \text{ and }
 \quad B_1'=
 \begin{pmatrix}
   1 & \alpha & 0\\
 0 & 1 & 0 \\
0 & 0& \gamma
 \end{pmatrix}$$ for some $\gamma>0$, the three-dimensional deformation would then be of the form $\varphi(x_1,x_2,x_3)=(u_1(x_1,x_2),u_2(x_1,x_2),\gamma x_3)$.

In thin films, however, a different relation between the two- and the three-dimensional problem is possible. Indeed, if the film is sufficiently thin, then incompatibility in the transverse direction leads to minor energy contributions. It is therefore reasonable to expect microstructure which is only compatible in-plane [@ChaudhuriMueller2006; @Hornung2008]. In this case, the two eigenstrains have the same determinant as $3\times 3$ matrices, but there is no reason for the $2\times 2$ blocks which determine the in-plane microstructure to have the same determinant, as for example with the three-dimensional eigenstrains $$A_2'= 
 \begin{pmatrix}
   1 & 0 & 0\\
 0 & 1-\alpha & 0 \\
0 & 0 &1/(1-\alpha)
 \end{pmatrix} \text{ and }
 \quad B_2'=
 \begin{pmatrix}
   1 & 0 & 0\\
 0 & 1+\alpha & 0 \\
0 & 0& 1/(1+\alpha)
 \end{pmatrix}\,.$$ It would be interesting to see if the difference between $K_1$ and $K_2$ results in a different scaling in the thin-film geometry studied in [@ChaudhuriMueller2006; @Hornung2008].

If no rank-one 