Discussion
Suppose N is the incidence matrix of a BIBD with parameters \((v, b, r, k, \lambda)\), then :
Answer: A
rank \((NN') = v\).
For a BIBD, every treatment occurs in \(r\) blocks and every pair together in \(\lambda\) blocks, so
\[ NN' = (r - \lambda)I_{v} + \lambda J_{v} \]
Its eigenvalues are \(rk\) (once) and \(r - \lambda\) (\(v - 1\) times), and \(r \gt \lambda\) in any proper BIBD, so the determinant \(rk(r - \lambda)^{v-1}\) is non-zero and \(NN'\) is a non-singular \(v \times v\) matrix of rank \(v\).
This is the proof of Fisher's inequality \(b \ge v\): rank \(N \ge\) rank \(NN' = v\). So the rank is \(b\) only for symmetric designs, and never \(v - 1\) or \(b - 1\).
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