 Solution

Let $B=A^{T}A$. It is a symmetric matrix, and ${\mathbf{v}}^{T}B\mathbf{v}=\Vert A\mathbf{v}{\Vert }^2\ge 0$. By the spectral theorem for real symmetric matrices, choose an orthonormal basis ${\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3$ of eigenvectors of $B$, with nonnegative eigenvalues ${\lambda }_1,{\lambda }_2,{\lambda }_3$.

For each $i\ne j$, take the unit parallelogram parametrized by $s{\mathbf{v}}_i+t{\mathbf{v}}_j$, $0\le s,t\le 1$. Its surface area is $1$. Moreover,The image parallelogram consequently has area $\sqrt{{\lambda }_i{\lambda }_j}$. By the assumption this equals $1$ for all three pairs, givingNo eigenvalue can be zero. Dividing the equations and using positivity yields ${\lambda }_1={\lambda }_2={\lambda }_3=1$. HenceThus testing all surfaces is stronger than necessary: the three unit parallelograms in these principal directions already determine the result. This is the three-dimensional characterization of linear surface-area isometries.