.* and * give me different answer
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
yoohooo's
am 28 Jul. 2019
Kommentiert: Stephen23
am 28 Jul. 2019
F = [1.5 0.2; 0.2 1.3];
I = eye(2);
E = 1/2*(F'*F-I);
E =
0.6450 0.2800
0.2800 0.3650
E = 1/2*(F'.*F-I);
E =
0.6250 0.0200
0.0200 0.3450
Why did * operates differ from .*? Can someone show the math behind the .* operator? When I was doing it by hand I got the answer same as * operator.
Akzeptierte Antwort
Weitere Antworten (0)
Siehe auch
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!