Your matrix is 4x5. How do you want to define a determinant for it ?

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

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% 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);

### Accepted Answer

Hassaan
on 9 May 2024

Edited: Hassaan
on 9 May 2024

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:');

disp(detA);

double(solve(detA==0,w,'MaxDegree',3))

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##### 5 Comments

Sam Chak
on 10 May 2024

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:');

disp(detA);

Wsol = double(solve(detA==0, W, 'MaxDegree', 3))

Torsten
on 10 May 2024

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.

### More Answers (2)

John D'Errico
on 9 May 2024

Edited: John D'Errico
on 9 May 2024

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

Assuming that is correctly your matrix, the result will be a degree 6 polynomial.

Adet = det(A)

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)

As you can see, there were no real solutions. All solutions were purely imaginary. The real parts of those solutions are all effectively zero.

##### 0 Comments

john
on 22 May 2024 at 13:37

Edited: john
on 23 May 2024 at 11:07

##### 0 Comments

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