How to combine/aggregate multiple Gram matrices to yield a one matrix by doing some desired matrix manipulations?

Let,

X = [1, 2, 3, 4; 
     5, 6, 7, 8;
     9, 10, 11, 12;
     13, 14, 15, 16;
     17, 18, 19, 20;
     21, 22, 23, 24];

% Class labels
Y = [a; a; b; b; c; c];

G = X'*X; % Size of G would be 6 x 6

Binary combinations of above classes (i.e., 3 choose 2 == 3) are {a, b}, {b, c}, {c, a}.

Matrices corresponding to above binary combinations would be:

X1 = 
[1, 2, 3, 4; 
5, 6, 7, 8; 
9, 10, 11, 12;
12, 14, 15, 16];

X2 =
9, 10, 11, 12;
13, 14, 15, 16;
17, 18, 19, 20;
21; 22, 23, 24];

X3 =
[17, 18, 19, 20;
 21, 22, 23, 24;
 1, 2, 3, 4; 
 5, 6, 7, 8;];

Gram matrices for X1, X2, X3 would be

G1 = X1'*X1; % size 4 x 4
G2 = X2'*X2; % size 4 x 4
G3 = X3'*X3; % size 4 x 4

Now suppose there is some z = [0.1, 0.2, 0.3];
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Questions:

I will like to multiple each Gram matrices by the elements of z, i.e.,

0.1 * G1;
0.2 * G2;
0.3 * G2;

and then combine them to give a Gram matrix equivalent to Gram matrix G which is of size 6 x 6.

So in other words, is there any way to multiple elements of Gram matrix G by the elements z such that

• 0.1 multiplies only those entries of G which correspond to the entries of Gram matrix G1
• 0.2 multiplies only those entries of G which correspond to the entries of Gram matrix G2
• 0.3 multiples only those entries of G which correspond to the entries of Gram matrix G3

I see two approaches to solve this problem:

1. Combine/Aggregate 0.1 * G1, 0.2 * G2, 0.3 * G3 (each of which are of size 4 x 4) to yield an output matrix of size 6 x 6 equivalent to Gram matrix G.
2. Directly multiple 0.1, 0.2, 0.3 with those elements of G which are corresponding to Gram matrices G1, G2, G3 respectively.

The first approach seems a lot of computation as it requires to compute three Gram matrices separately. Besides, the main problem here is that I do not know how to combine/aggregate those Gram matrices to yield one of size 6 x 6 as if it were obtained by original data matrix G = X'*X;

The second approach will take less computation, but again I do not know how to obtain the desired result.

One final note: Here, I took z with only three elements [0.1, 0.2, 0.3] because there are three classes {a, b, c}. So as a general solution, z contains as many elements as class numbers.

I will appreciate any advise!