= Solution
A row's squared <Euclidean norm> is $[(1+\sqrt3)^2+1+(1-\sqrt3)^2]/6=3/2$, and the <dot product> of the two rows is $[2(1-3)+1]/6=-1/2$. Thus the row <Gram matrix> is
$$
\boxed{AA^{\mathsf T}=\begin{pmatrix}3/2&-1/2\\-1/2&3/2\end{pmatrix}.}
$$
For the column <Gram matrix>, the first and third squared column norms are $4/3$, the middle one is $1/3$, and the respective cross-products are $1/3,-2/3,1/3$. Hence
$$
\boxed{A^{\mathsf T}A=\frac13\begin{pmatrix}4&1&-2\\1&1&1\\-2&1&4\end{pmatrix}.}
$$
The two products are square <matrices> of different sizes; their unequal dimensions are consistent with $A$ being rectangular.
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