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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 

Π$_{jrv}$ = δ$_{jv}$. 

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 

and 

I js$_{ri}$v =      1 An individual on sequence s$_{ri}$ receives treatment j in period v, 0 otherwise . 

T$_{js}$$_{ri}$$_{v}$ = δ$_{1}$$_{I}$$_{js}$$_{ri}$$_{v}$ , 

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for all j and k . Therefore, we can determine the form of each of the scalar elements in the matrix above 

