i am tring to input different z values to get different B using matrix manipulation. just having trouble using for loop

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kaixi gu
kaixi gu 2023-2-1
编辑： Stephen23 2023-2-1
for z= 0.01:0.1:6
A = [-2.7/sqrt(20.25+z^2),3.2/sqrt(20.25+z^2),z/sqrt(20.25+z^2);3.2/sqrt(10.24+z^2),0,z/sqrt(10.24+z^2);-2.7/sqrt(23.29+z^2),4/sqrt(23.29+z^2),z/sqrt(23.29+z^2)];
c = [0;0;10000];
B = inv(A)*c
end

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采纳的回答

Stephen23
Stephen23 2023-2-1
编辑：Stephen23 2023-2-1
Ran in:
Note that I replaced the INV()* with the recommended MLDIVIDE:
https://www.mathworks.com/help/matlab/ref/inv.html#bu8tv8c-6
Anyone who has read your code and the INV() documentation will make this recommendation.
c = [0;0;10000];
V = 0.01:0.1:6;
N = numel(V);
B = nan(numel(c),N);
for k = 1:N
z = V(k);
A = [-2.7/sqrt(20.25+z^2),3.2/sqrt(20.25+z^2),z/sqrt(20.25+z^2);3.2/sqrt(10.24+z^2),0,z/sqrt(10.24+z^2);-2.7/sqrt(23.29+z^2),4/sqrt(23.29+z^2),z/sqrt(23.29+z^2)];
B(:,k) = A\c; % recommended algorithm
end
B
B = 3×60
1.0e+07 * 0.0033 0.0033 0.0033 0.0033 0.0033 0.0033 0.0033 0.0033 0.0033 0.0033 0.0033 0.0034 0.0034 0.0034 0.0034 0.0034 0.0034 0.0035 0.0035 0.0035 0.0035 0.0036 0.0036 0.0036 0.0037 0.0037 0.0037 0.0038 0.0038 0.0038 0.0060 0.0060 0.0060 0.0060 0.0061 0.0061 0.0061 0.0061 0.0061 0.0061 0.0062 0.0062 0.0062 0.0063 0.0063 0.0063 0.0064 0.0064 0.0064 0.0065 0.0065 0.0066 0.0066 0.0067 0.0067 0.0068 0.0069 0.0069 0.0070 0.0070 -1.0470 -0.0952 -0.0499 -0.0338 -0.0256 -0.0206 -0.0173 -0.0149 -0.0131 -0.0117 -0.0106 -0.0097 -0.0089 -0.0083 -0.0077 -0.0073 -0.0069 -0.0065 -0.0062 -0.0059 -0.0056 -0.0054 -0.0052 -0.0050 -0.0049 -0.0047 -0.0046 -0.0044 -0.0043 -0.0042

更多回答（2 个）

Kunal Kandhari
Kunal Kandhari 2023-2-1
Code seems to be working!
Can you please elaborate what error you're getting?
  1 个评论
kaixi gu
kaixi gu 2023-2-1
the z should be a set of input value go into the matices. i am tring to get different 1*3 B as an output respect to the different input z. i just dont know how to use the for loop to get it.

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Tushar Behera
Tushar Behera 2023-2-1
编辑：Tushar Behera 2023-2-1
Hi kaixi,
I am assuming you want to get different B values for different values of Z, and want to keep all the B values. However the code you have written will give B value for the last value of Z.
in order to solve this issue you can create B as an array and save all the instances of the solution inside B.
For example
z= 0.01:0.1:6
num=numel(z)
B=cell(num,1)
i=1;
for z= 0.01:0.1:6
A = [-2.7/sqrt(20.25+z^2),3.2/sqrt(20.25+z^2),z/sqrt(20.25+z^2);3.2/sqrt(10.24+z^2),0,z/sqrt(10.24+z^2);-2.7/sqrt(23.29+z^2),4/sqrt(23.29+z^2),z/sqrt(23.29+z^2)];
c = [0;0;10000];
answer= inv(A)*c
B{i}=answer;
i=i+1;
end
You can change the code as per your requirements. I hope this resolves your query.
Regards,
Tushar

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