A quicker route is dimensional/rank bookkeeping instead of re-deriving the Lagrange-multiplier algebra: simply check which candidate expression is even matrix-conformable and produces the correct output shape \(3\times2\) (so that \(m_{3\times1} = [\,\cdot\,]_{3\times2}\,d_{2\times1}\)).
\(G\) is \(2\times3\), so \(G^T\) is \(3\times2\).
- (A) \((G^TG)^{-1}G^T\): \(G^TG\) is \(3\times3\) but has rank at most 2 (since \(G\) only has 2 independent rows) -- it is SINGULAR and cannot be inverted. Discard.
- (B) \(G^T(GG^T)^{-1}\): \(GG^T\) is \(2\times2\) and generically full rank 2 (invertible) when \(G\) has full row rank. The product \(G^T_{3\times2}\,(GG^T)^{-1}_{2\times2}\) is \(3\times2\) -- exactly the shape needed to map \(d_{2\times1}\to m_{3\times1}\). Valid and well-posed.
- (C) \((GG^T)^{-1}G\): \((GG^T)^{-1}_{2\times2}\,G_{2\times3}\) gives a \(2\times3\) matrix, which cannot even be multiplied by \(d_{2\times1}\) to give a \(3\times1\) model vector. Dimensionally wrong.
- (D) \(GG^TG^{-1}\): requires \(G^{-1}\), but \(G\) is rectangular (\(2\times3\)) and has no ordinary inverse at all. Meaningless.
Only option (B) survives both the invertibility check and the shape/dimension check, confirming it is the correct generalized inverse for this underdetermined (more unknowns than data) system -- exactly the minimum-norm inverse \(G^T(GG^T)^{-1}\).
\(\boxed{\text{Answer: (B)}}\)