Matrix multiplication using a for loop
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I want to make matrix A using elements of matrices P1 and P2 in the following form. Is there any way to do this using a for loop. Any help would be appreciated!
A(1,:) = [P1(1,1).*P2(1,1:4) P1(1,2).*P2(1,1:4) P1(1,3).*P2(1,1:4) P1(1,4).*P2(1,1:4)]
A(2,:) = [P1(2,1).*P2(2,5:8) P1(2,2).*P2(2,5:8) P1(2,3).*P2(2,5:8) P1(2,4).*P2(2,5:8)]
A(3,:) = [P1(3,1).*P2(3,9:12) P1(3,2).*P2(3,9:12) P1(3,3).*P2(3,9:12) P1(3,4).*P2(3,9:12)]
A(4,:) = [P1(4,1).*P2(4,13:16) P1(4,2).*P2(4,13:16) P1(4,3).*P2(4,13:16) P1(4,4).*P2(3,13:16)]
A(5,:) = [P1(5,5).*P2(5,1:4) P1(5,6).*P2(5,1:4) P1(5,7).*P2(5,1:4) P1(5,8).*P2(5,1:4)]
A(6,:) = [P1(6,5).*P2(6,5:8) P1(6,6).*P2(6,5:8) P1(6,7).*P2(6,5:8) P1(6,8).*P2(5,5:8)]
A(7,:) = [P1(7,5).*P2(7,9:12) P1(7,6).*P2(7,9:12) P1(7,7).*P2(7,9:12) P1(7,8).*P2(7,9:12)]
A(8,:) = [P1(8,5).*P2(8,13:16) P1(8,6).*P2(8,13:16) P1(8,7).*P2(8,13:16) P1(8,8).*P2(8,13:16)]
A(9,:) = [P1(9,9).*P2(9,1:4) P1(9,10).*P2(9,1:4) P1(9,11).*P2(9,1:4) P1(9,12).*P2(9,1:4)]
A(10,:) = [P1(10,9).*P2(10,5:8) P1(10,10).*P2(10,5:8) P1(10,11).*P2(10,5:8) P1(10,12).*P2(10,5:8)]
A(11,:) = [P1(11,9).*P2(11,9:12) P1(11,10).*P2(11,9:12) P1(11,11).*P2(11,9:12) P1(11,12).*P2(11,9:12)]
A(12,:) = [P1(12,9).*P2(12,13:16) P1(12,10).*P2(12,13:16) P1(12,11).*P2(12,13:16) P1(12,12).*P2(12,13:16)]
A(13,:) = [P1(13,13).*P2(13,1:4) P1(13,14).*P2(13,1:4) P1(13,15).*P2(13,1:4) P1(13,16).*P2(13,1:4)]
A(14,:) = [P1(14,13).*P2(14,5:8) P1(14,14).*P2(14,5:8) P1(14,15).*P2(14,5:8) P1(14,16).*P2(14,5:8)]
A(15,:) = [P1(15,13).*P2(15,9:12) P1(15,14).*P2(15,9:12) P1(15,15).*P2(15,9:12) P1(15,16).*P2(15,9:12)]
A(16,:) = [P1(16,13).*P2(16,13:16) P1(16,14).*P2(16,13:16) P1(16,15).*P2(16,13:16) P1(16,16).*P2(16,13:16)]
1 commentaire
Wilmer Alexander - Conferencista en Automatización y Robótica más influyente en Iberoamérica
le 28 Juil 2020
Modifié(e) : Wilmer Alexander - Conferencista en Automatización y Robótica más influyente en Iberoamérica
le 28 Juil 2020
Thanks for asking. If possible. I remain at your disposal for any questions.
Gracias por preguntar. Si es posible. Quedo a su disposición para cualquier consulta.
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