How to add unknow parameter in matrix and solve it by use det() syntax for finding w

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Trong Nhan Tran
Trong Nhan Tran 2024-5-9
编辑： john 2024-5-23
% under is what i did but seen it is not work for det(A) for find w
clc % clear history command and past result
syms w;
m1 = 1.8;
m2 = 6.3;
m3 = 5.4;
m4 = 22.5;
m5 = 54;
c2 = 10000;
c3 = 500;
c4 = 1500;
c5 = 1100;
k2 = 1*10^8;
k3 = 50*10^3;
k4 = 75*10^3;
k5 = 10*10^3;
% Form of matrix is Ax=b
% Where A is nxn matrix, x is displacement of lumped masses and b is RHS.
A= [0, 0, 0, 0, (m5*w^2)-k5-c5;
0, 0, k4+c4, -k4-c4+(m4*w^2)+k5+c5, -k5+c5;
k2+c2, -k3-c3-k2-c2+(m2*w^2), k3+c3, 0, 0;
-k2-c2+(m1*w^2), k2+c2, 0, 0, 0];
det (A);

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Hassaan
Hassaan 2024-5-9
编辑：Hassaan 2024-5-9
Ran in:
clc; % Clear command window
clear; % Clear workspace
syms w; % Define w as a symbolic variable
% Define masses, damping coefficients, and stiffness coefficients
m1 = 1.8; m2 = 6.3; m3 = 5.4; m4 = 22.5; m5 = 54;
c2 = 10000; c3 = 500; c4 = 1500; c5 = 1100;
k2 = 1*10^8; k3 = 50*10^3; k4 = 75*10^3; k5 = 10*10^3;
% Define the matrix A
A = [k2+c2, -k2-c2+(m2*w^2), 0, 0, 0;
-k2-c2, k2+c2+k3+c3, -k3-c3, 0, 0;
0, -k3-c3, k3+c3+k4+c4, -k4-c4+(m4*w^2), 0;
0, 0, -k4-c4, k4+c4+k5+c5, -k5-c5;
0, 0, 0, -k5, k5+c5+(m5*w^2)];
% Calculate the determinant of the matrix A
detA = det(A);
% Display the determinant
disp('The determinant of matrix A is:');
The determinant of matrix A is:
disp(detA);
double(solve(detA==0,w,'MaxDegree',3))
ans =
1.0e+02 * 0.0000 + 0.0369i 0.0000 + 1.0353i 0.0000 - 0.2352i -0.0000 - 0.0369i -0.0000 - 1.0353i -0.0000 + 0.2352i
-----------------------------------------------------------------------------------------------------------------------------------------------------
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It's important to note that the advice and code are based on limited information and meant for educational purposes. Users should verify and adapt the code to their specific needs, ensuring compatibility and adherence to ethical standards.
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  5 个评论
显示 3更早的评论
Sam Chak
Sam Chak 2024-5-10
Ran in:
@Trong Nhan Tran,
Could you explain what the symbolic variable w is?
syms w W
% Define masses, damping coefficients, and stiffness coefficients
m1 = 1.8; m2 = 6.3; m3 = 5.4; m4 = 22.5; m5 = 54;
c2 = 10000; c3 = 500; c4 = 1500; c5 = 1100;
k2 = 1*10^8; k3 = 50*10^3; k4 = 75*10^3; k5 = 10*10^3;
% Define the matrix A
A = [k2+c2, -k2-c2+(m2*w^2), 0, 0, 0;
-k2-c2, k2+c2+k3+c3, -k3-c3, 0, 0;
0, -k3-c3, k3+c3+k4+c4, -k4-c4+(m4*w^2), 0;
0, 0, -k4-c4, k4+c4+k5+c5, -k5-c5;
0, 0, 0, -k5, k5+c5+(m5*w^2)];
% Calculate the determinant of the matrix A
detA = det(A);
detA = subs(detA, w^2, W);
% Display the determinant
disp('The determinant of matrix A is:');
The determinant of matrix A is:
disp(detA);
Wsol = double(solve(detA==0, W, 'MaxDegree', 3))
Wsol =
1.0e+04 * -0.0014 + 0.0000i -1.0718 + 0.0000i -0.0553 - 0.0000i
Torsten
Torsten 2024-5-10
@Trong Nhan Tran
I dont know why but when i use det(A) the error is Matrix must be square.
Maybe you used the 4x5 matrix you posted first.

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更多回答（2 个）

John D'Errico
John D'Errico 2024-5-9
编辑：John D'Errico 2024-5-9
Ran in:
syms w;
m1 = 1.8;
m2 = 6.3;
m3 = 5.4;
m4 = 22.5;
m5 = 54;
c2 = 10000;
c3 = 500;
c4 = 1500;
c5 = 1100;
k2 = 1*10^8;
k3 = 50*10^3;
k4 = 75*10^3;
k5 = 10*10^3;
% Form of matrix is Ax=b
% Where A is nxn matrix, x is displacement of lumped masses and b is RHS.
A = [k2+c2, -k2-c2+(m2*w^2), 0, 0, 0;
-k2-c2, k2+c2+k3+c3, -k3-c3, 0, 0;
0, -k3-c3, k3+c3+k4+c4, -k4-c4+(m4*w^2), 0;
0, 0, -k4-c4, k4+c4+k5+c5, -k5-c5;
0, 0, 0, -k5, k5+c5+(m5*w^2)];
A
A = 
Assuming that is correctly your matrix, the result will be a degree 6 polynomial.
Adet = det(A)
Adet = 
There can be no exact algebraic solutions fro a degree 5 or higher polynomial. But you can have numerically computed roots.
wsol = solve(Adet,maxdegree = 6)
wsol = 
As you can see, there were no real solutions. All solutions were purely imaginary. The real parts of those solutions are all effectively zero.
john
john 2024-5-22
编辑：john 2024-5-23
To add an unknown parameter in a matrix and solve it using det() syntax for an IQ brain test, replace an element with 'w.' Then, compute the determinant and solve the resulting equation for 'w.' This process allows for a more dynamic and challenging matrix calculation, enhancing the complexity of the IQ brain test.

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