Effacer les filtres
Effacer les filtres

multiply a matrix by a specific value from a vector

3 vues (au cours des 30 derniers jours)
Naema
Naema le 8 Avr 2015
Modifié(e) : Naema le 8 Avr 2015
Hi : I want to multiply a matrix (resultx) sized 10001x10001 by only one value from etae which is sized 1x50.I want to access only the first value of etae and the last value of eate separately. can I do it like this:
Iph1=pinc*(resultx).^2*(etae(1,1))/h*v) % for first value
and,
Iph2=pinc*(resultx).^2*(etae(1,50))/h*v) % for last value
and, did I square (resultx) correctly?. should the (.) be there?. pinc,h, v, etae are single values. resultx is (mXn)
  2 commentaires
James Tursa
James Tursa le 8 Avr 2015
What are the sizes of pinc, h, and v? Scalars? As written, only h is in the denominator. Is this what you want or do you want v there as well?
Naema
Naema le 8 Avr 2015
Modifié(e) : Naema le 8 Avr 2015
both h and v are in the denominator. pinc,h,v are scalars

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Jeffrey Girard
Jeffrey Girard le 8 Avr 2015
Modifié(e) : Jeffrey Girard le 8 Avr 2015
Let's use a cleaner example:
a = 5;
b = [1,2,3,4;5,6,7,8];
c = [0.5,1.0,1.5,2.0];
So if I want to multiply the matrix b by scalar a and then add that product's square to the first value of vector c, you should do the following:
d = (b .* a) .^ 2 + c(1);
And if you want to do the same but with the last value of vector c , you should do the following:
d = (b .* a) .^ 2 + c(end);
The dot before a mathematical operator (e.g., .*) indicates that you want to perform the operation using array operations (or element-wise) rather than matrix operations. If you are unfamiliar with matrix algebra, you want to be using the dots (i.e., array operations). In some cases, such as when using scalars, the two are equivalent, but it will be safer for you to use the dots as I did above.
  1 commentaire
Naema
Naema le 8 Avr 2015
Modifié(e) : Naema le 8 Avr 2015
so, is this the right answer?:
Iph1=pinc.*(resultx).^2.*etae(1)/h/v;

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