If your matrix can have a number of rows that is a multiple of three, and you should know its size at the time you need to perform these calculations, then you can use reshape and then rely on the behavior of the mean function when operating on matrices like this:
>> A = rand(5,6)
A =
0.8147 0.0975 0.1576 0.1419 0.6557 0.7577
0.9058 0.2785 0.9706 0.4218 0.0357 0.7431
0.1270 0.5469 0.9572 0.9157 0.8491 0.3922
0.9134 0.9575 0.4854 0.7922 0.9340 0.6555
0.6324 0.9649 0.8003 0.9595 0.6787 0.1712
>> A = rand(15,6)
A =
0.7060 0.4898 0.7513 0.3500 0.0759 0.6020
0.0318 0.4456 0.2551 0.1966 0.0540 0.2630
0.2769 0.6463 0.5060 0.2511 0.5308 0.6541
0.0462 0.7094 0.6991 0.6160 0.7792 0.6892
0.0971 0.7547 0.8909 0.4733 0.9340 0.7482
0.8235 0.2760 0.9593 0.3517 0.1299 0.4505
0.6948 0.6797 0.5472 0.8308 0.5688 0.0838
0.3171 0.6551 0.1386 0.5853 0.4694 0.2290
0.9502 0.1626 0.1493 0.5497 0.0119 0.9133
0.0344 0.1190 0.2575 0.9172 0.3371 0.1524
0.4387 0.4984 0.8407 0.2858 0.1622 0.8258
0.3816 0.9597 0.2543 0.7572 0.7943 0.5383
0.7655 0.3404 0.8143 0.7537 0.3112 0.9961
0.7952 0.5853 0.2435 0.3804 0.5285 0.0782
0.1869 0.2238 0.9293 0.5678 0.1656 0.4427
>> Acolumn2 = A(:,2)
Acolumn2 =
0.4898
0.4456
0.6463
0.7094
0.7547
0.2760
0.6797
0.6551
0.1626
0.1190
0.4984
0.9597
0.3404
0.5853
0.2238
Areshape = reshape(Acolumn2,3,5)
Areshape =
0.4898 0.7094 0.6797 0.1190 0.3404
0.4456 0.7547 0.6551 0.4984 0.5853
0.6463 0.2760 0.1626 0.9597 0.2238
>> mean(Areshape)
ans =
0.5272 0.5800 0.4991 0.5257 0.3832
