How can I multiply each row of 3 matrices individually?

Hello everyone,

I have 3 matrixes to be multiplied. x[36,36],r[36,36],a[36x1].

What i wanna do is this;

x(1,1)*r(1,1)*ai(1)+x(2,1)*r(2,1)*ai(2)+....+x(35,1)*r(35,1)*ai(35)+x(36,1)*r(36,1)*ai(36) <=Kj*y(1)

x(1,2)*r(1,2)*ai(1)+x(2,2)*r(2,2)*ai(2)+....+x(35,2)*r(35,2)*ai(35)+x(36,2)*r(36,2)*ai(36)<=Kj*y(2)

...

x(1,36)*r(1,36)*ai(1)+x(2,36)*r(2,36)*ai(2)+....+x(35,36)*r(35,36)*ai(35)+x(36,36)*r(36,36)*ai(36)<=Kj*y(36)

This is what i wrote so far;

sum(x.*rij*ai,2)- Kj*y(:,1)<=0

this does what i want to do with the right side( Kj*y()) but the left side is the opposite what i want.

it does this;

x(1,1)*r(1,1)*ai(1)+x(1,2)*r(1,2)*ai(2)+....+x(1,35)*r(1,35)*ai(35)+x(1,36)*r(1,36)*ai(36)<=Kj*y(1)

hope this is clear. open to every suggestions. What i am looking for is only one line! Like the one i tried writing!

Thank you in advance!

Answer 1

[David Hill]

sum(x.*rij*ai)- Kj*y(:,1)<=0;%want to add columns (delete ,2)

[Azime Beyza Ari]

where should i write (delete ,2) ?

[David Hill]

sum(x.*rij*ai)

[Azime Beyza Ari]

Oh, sorry!

Already tried that. But what i want is row multpilication only. i have 36 rows for x and r. but when i do this ;

sum(x.*rij*ai)

it just multiplies and sums all of the matrix. And it does this for 36 times. So not what i am looking for.

Answer 2

[Star Strider]

The part of this involving K and y is ambiguous.

It would help to know what and y are, because it appears that Kis a vector, and y is a matrix, the columns of which are used to multiply K to test the inequality.

x = randi(9,4)

x = 4×4

1 9 6 3 6 5 3 2 7 9 9 6 8 8 9 6

r = randi(9,4)

r = 4×4

7 8 6 9 6 9 5 9 9 8 6 9 2 3 5 4

a = randi(9,4,1)

a = 4×1

9 9 7 4

z = (x .* r)' .* a

z = 4×4

63 324 567 144 648 405 648 216 252 105 378 315 108 72 216 96

z_sum = sum(z,2)

z_sum = 4×1

1098 1917 1050 492

K = 92;
y = randi(9,1,4)

y = 1×4

9 2 6 4

Ky = K .* y

Ky = 1×4

828 184 552 368

z_logical = z_sum <= Ky

z_logical = 4×4 logical array
   0   0   0   0
   0   0   0   0
   0   0   0   0
   1   0   1   0

z_logical = sum((x .* r)' .* a, 2) <= Ky                        % Single-Line Version

z_logical = 4×4 logical array
   0   0   0   0
   0   0   0   0
   0   0   0   0
   1   0   1   0

for k1 = 1:size(x,1)                                            % This Checks To Be Certain That The 'z' Matrix Is Caclulated Correctly
    for k2 = 1:size(x,2)
        z(k2,k1) = x(k2,k1) * r(k2,k1) * a(k2);                 % Check
    end
end
z

z = 4×4

63 648 324 243 324 405 135 162 441 504 378 378 64 96 180 96

EDIT — (11 Apr 2022 at 17:48)

Changed ‘K’ to be a constant scalar, and ‘y’ to be a row vector.

Since ‘z_sum’ is a column vector, the result of the logical comparison with ‘Ky’ will be a logical matrix. If optimtool wants a scalar result, it will be necessary to do further processing on ‘z_logical’.

.

[Azime Beyza Ari]

sorry for unclear information. K is a 92. y has dimensions (1x36).

Also thank you for your answer. But what i am looking for is just only one line. I am using optimtool and trying to write it on only one line.

Answer 3

[Torsten]

(x.*rij).' * ai - Kj*y <= 0

assuming that ai and y are column vectors.

[Azime Beyza Ari]

Yeah did not worked. Gives error saying; Incorrect dimensions for matrix multiplication. check if the no of columns in first matrixx matches the no of rows in the second matrix.

but they match.

[Torsten]

Is y a vector or a matrix ?

Because you wrote y(:,1) instead of y above.

Or do you have to specify the "0" of the right-hand side as "zeros(size(y,1),1)" ?

Try also

(x.*rij)'*ai - Kj*y <= 0