= Solution
A visibly different primitive parallelogram is generated by
$$
\boxed{\mathbf a'_1=\mathbf a_1=a(1,1),\qquad
\mathbf a'_2=\mathbf a_1+\mathbf a_2=a(0,2)}.
$$
Its vertices, relative to one lattice point, are $(0,0)$, $(a,a)$, $(a,3a)$ and $(0,2a)$. Its area is
$$
\boxed{A'_c=\left|\det\begin{pmatrix}a&0\\a&2a\end{pmatrix}\right|=2a^2}.
$$
The basis change has integer matrix $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ with determinant one. It is invertible over the integers, so the new vectors generate every original lattice point, not merely a sublattice. The cell is therefore primitive despite being sheared relative to cell B.
\Image[/past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2023/paper-2-primitive-cells.png]
{title=Two primitive bases generate the same centered-square lattice}
{description=Original coordinate construction from the derived translation vectors, with axes measured in units of $a$. The left cell uses the diagonal orthogonal basis; the right uses a determinant-one shear. Both highlighted parallelograms have area $2a^2$ and one lattice point per cell.}
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