How do I solve this matrix in a loop for different values for P?

x1= [-0.005:0.005]

x2= [-0.005:0.005]

x3= [-0.005:0.005]

x4= [-0.005:0.005]

x5= [-0.005:0.005]

w1= [-0.005:0.005]

w2= [-0.005:0.005]

w3= [-0.005:0.005]

w4= [-0.005:0.005]

w5= [-0.005:0.005]

P=[0 x1 w1

0 x2 w2

0 x3 w3

0 x4 w4

0 x5 w5]

A=[1.0000 0.005 0.02

1.0000 0.014 0.17

1.0000 0.022 0.42

1.0000 0.036 1.16

1.0000 0.052 2.44]

Y= A+P

B=[0.86

0.76

0.74

0.57

0.39]

aa=Y\B

[Rik]

What exactly is your question? Which different values for P would you want to have? Your code already runs.

x1= [-0.005:0.005];
x2= [-0.005:0.005];
x3= [-0.005:0.005];
x4= [-0.005:0.005];
x5= [-0.005:0.005];
w1= [-0.005:0.005];
w2= [-0.005:0.005];
w3= [-0.005:0.005];
w4= [-0.005:0.005];
w5= [-0.005:0.005];
P=[0 x1	w1
   0 x2 w2
   0 x3 w3
   0 x4 w4
   0 x5 w5];
A=[1.0000 0.005 0.02		
   1.0000 0.014 0.17
   1.0000 0.022 0.42
   1.0000 0.036 1.16
   1.0000 0.052 2.44];
Y= A+P;
B=[0.86				
   0.76
   0.74
   0.57
   0.39];
aa=Y\B

aa = 3×1

0.8534 -6.4035 -0.0672

[Jan]

The question is not clear. Simply run the same code with the different values of P.

Some hints:

[] is Matlab's operator for the concatenation of arrays. With [-0.005:0.005] you contcate the vector -0.005:0.005 with nothing, therefore the square brackets are a waste of time only.

[ABHISHEK JHA]

Thanks for the reply. How could I define P so that it takes different values of x1,x2... w1,w2?

[ABHISHEK JHA]

I want to run this code for error analysis by slightly changing the Values of A with the help of Matrix P. I am having trouble choosing a random set of values of P and collecting different set of values

Answer 1

[Rik]

I'm guessing you want a for loop like this:

x1= linspace(-0.005,0.005,10);
x2= linspace(-0.005,0.005,10);
x3= linspace(-0.005,0.005,10);
x4= linspace(-0.005,0.005,10);
x5= linspace(-0.005,0.005,10);
w1= linspace(-0.005,0.005,10);
w2= linspace(-0.005,0.005,10);
w3= linspace(-0.005,0.005,10);
w4= linspace(-0.005,0.005,10);
w5= linspace(-0.005,0.005,10);
A=[1.0000 0.005 0.02		
   1.0000 0.014 0.17
   1.0000 0.022 0.42
   1.0000 0.036 1.16
   1.0000 0.052 2.44];
B=[0.86				
   0.76
   0.74
   0.57
   0.39];
aa=zeros(size(A,2),numel(x1));
for n=1:numel(x1)
    P=[0 x1(n) w1(n)
        0 x2(n) w2(n)
        0 x3(n) w3(n)
        0 x4(n) w4(n)
        0 x5(n) w5(n)];
    Y= A+P;
    aa(:,n)=Y\B;
end
aa

aa = 3×10

0.8534 0.8606 0.8678 0.8750 0.8822 0.8894 0.8966 0.9038 0.9109 0.9181 -6.4035 -6.4035 -6.4035 -6.4035 -6.4035 -6.4035 -6.4035 -6.4035 -6.4035 -6.4035 -0.0672 -0.0672 -0.0672 -0.0672 -0.0672 -0.0672 -0.0672 -0.0672 -0.0672 -0.0672

[ABHISHEK JHA]

Thanks for the quick help.Appreciate it!!