multiply a matrix by a specific value from a vector

2 visualizzazioni (ultimi 30 giorni)
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 Commenti
James Tursa
James Tursa il 8 Apr 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 il 8 Apr 2015
Modificato: Naema il 8 Apr 2015
both h and v are in the denominator. pinc,h,v are scalars

Accedi per commentare.

Risposta accettata

Jeffrey Girard
Jeffrey Girard il 8 Apr 2015
Modificato: Jeffrey Girard il 8 Apr 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 Commento
Naema
Naema il 8 Apr 2015
Modificato: Naema il 8 Apr 2015
so, is this the right answer?:
Iph1=pinc.*(resultx).^2.*etae(1)/h/v;

Accedi per commentare.

Più risposte (0)

Categorie

Scopri di più su Operators and Elementary Operations in Help Center e File Exchange

Tag

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by