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

% 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);

[Torsten]

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

[Trong Nhan Tran]

sry i missed one comma when typing

Answer 1

[Hassaan]

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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[Torsten]

... and now use

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

[Hassaan]

@Torsten Updated my answer. Thank you.

[Trong Nhan Tran]

I dont know why but when i use det(A) the error is Matrix must be square.

[Sam Chak]

@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]

@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.

Answer 2

[John D'Errico]

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.

Answer 3

[john]

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.