Here, dim( Π$_{jr}$ ) = dim( 1$_{r}$ ) = dim( T$_{js}$$_{r}$$_{i}$ ) = r × 1, for all i and j . Therefore 1 T $_{r}$Σ − 1 r 1$_{r}$ , Π T $_{jr}$Σ − 1 r 1$_{r}$ , T T $_{js$_{ri}$}$Σ − 1 r 1$_{r}$ , 1 T $_{r}$Σ − 1 r Π$_{jr}$ , 1 T $_{r}$Σ − 1 r T$_{js}$$_{ri}$ , Π T $_{jr}$Σ − 1 r Π$_{kr}$ , T T $_{js$_{ri}$}$Σ − 1 r T$_{ks}$$_{ri}$ and Π T $_{jr}$Σ − 1 r T$_{ks}$$_{ri}$ are scalars for all i , j and k . By the definition of being in period j , the v th element of Π$_{jr}$ is given by Whilst, since complete block design sequences are used, the v th element of T$_{js}$$_{ri}$ is given by where We denote this non-zero element, if it exists, by t$_{j}$ . Now from the symmetry present in Σ − 1 r , using Lemma 4.1 we have for all j and k . Therefore, we can determine the form of each of the scalar elements in the matrix above and X T $_{s$_{ri}$}$Σ − 1 r X$_{s}$$_{ri}$ =        1 T r Π T 2 r . . T T D − 1 s$_{ri}$        Σ − 1 r $^{(}$1$_{r}$ Π$_{2}$$_{r}$ . . . T$_{D}$$_{−}$$_{1}$$_{s}$$_{ri}$ ) =        1 T $_{r}$Σ − 1 r 1$_{r}$ 1 T $_{r}$Σ − 1 r Π$_{2}$$_{r}$ . . . 1 T $_{r}$Σ − 1 r T$_{D}$$_{−}$$_{1}$$_{s}$$_{ri}$ Π T 2 $_{r}$Σ − 1 r 1$_{r}$ Π T 2 $_{r}$Σ − 1 r Π$_{2}$$_{r}$ . . . Π T 2 $_{r}$Σ − 1 r T$_{D}$$_{−}$$_{1}$$_{s}$$_{ri}$ . . . . . . . . . T T D − 1 $_{s$_{ri}$}$Σ − 1 r 1$_{r}$ T T D − 1 $_{s$_{ri}$}$Σ − 1 r Π$_{2}$$_{r}$ . . . T T D − 1 $_{s$_{ri}$}$Σ − 1 r T$_{D}$$_{−}$$_{1}$$_{s}$$_{ri}$$^{}$       . I js$_{ri}$v =      1 An individual on sequence s$_{ri}$ receives treatment j in period v, 0 otherwise . ∑ j,k Σ − 1 rjk = rσ 2 e σ 2 $_{e}$( σ 2 e + rσ 2 $_{b}$) , ∑ j Σ − 1 rjk = ∑ k Σ − 1 rjk = σ 2 e σ 2 $_{e}$( σ 2 e + rσ 2 $_{b}$) , Π$_{jrv}$ = δ$_{jv}$. T$_{js}$$_{ri}$$_{v}$ = δ$_{1}$$_{I}$$_{js}$$_{ri}$$_{v}$ , 21