how to develop n order matrix?

i wish to make a matrix of nth order fromm K=[2000,-1000,0,0;-1000,2000,-1000,0;0,-1000,2000,-1000;0,0,-1000,1000; so on to nth order]

can anyone please help me with this ...

[Jan]

The explanation is not clear yet. What do you call "n.th order"?

It is getting a liitle bit easier to guess, if you post the matrix in 2D:

K = [2000, -1000, 0,    0; ...
    -1000, 2000, -1000, 0; ...
    0,    -1000, 2000, -1000; ...
    0,     0,   -1000,  1000;    

But why is the last element 1000 and not 2000?

[Jasneet Singh]

its because last element is not hinged or fixed. so, there is only 1 force=1000 acting on it.

[Walter Roberson]

K=[2000,-1000,0,0;-1000,2000,-1000,0;0,-1000,2000,-1000;0,0,-1000,1000]

K = 4×4

2000 -1000 0 0 -1000 2000 -1000 0 0 -1000 2000 -1000 0 0 -1000 1000

Answer 1

[Jan]

With some guessing: You want a tridiagonal matrix with the right bottom element changed. Then:

n = 4;
K = diag(repmat(2000, 1, n)) + ...
    diag(repmat(-1000, 1, n-1), 1) + ...
    diag(repmat(-1000, 1, n-1), -1)

or

K  = zeros(n, n);
nn = n * n;
n1 = n + 1;
K( 1:n1:nn)   =  2000;
K(n1:n1:nn)   = -1000;
K( 2:n1:nn-n) = -1000;

or

K = toeplitz([2000, -1000, zeros(1, n - 2)])

or

K = full(gallery('tridiag', n, -1000, 2000, -1000))

or

K = conv2(eye(n), [-1000, 2000, -1000], 'same')

any finally:

K(n, n) = 1000;

Or directly:

K = diag([repmat(2000, 1, n - 1), 1000]) + ...   % Last element adjusted
    diag(repmat(-1000, 1, n - 1), 1) + ...
    diag(repmat(-1000, 1, n - 1), -1)

[Jasneet Singh]

can you please specify how you did this ?

[Jan]

How I did what?

I've added changing the last element to 1000.

[Jasneet Singh]

i mean from where can i learn this codes from to make my life easier? like one you have used recenty for coding..

[Jan]

In your case it helps to recognize, that you want almost a "tridiagonal matrix". Then asking an internet search engine for "Matlab tridiagonal" finds matching code snippets. But if you do not know the term "tridiagonal", it is really hard to find matching keywpord. Then asking here in the forum is a very good method to let others find the keywords and to learn MATLAB. This is the purpose of this forum.

[Jasneet Singh]

for omegaf_4=[0:1:80]       % forcing frequency w4 form 1:80 with a resolution of 1Hz
    
    F_b=[0;0;1;0]; 
   
    D4=(-(omegaf_4^2).*M+K+(i*omegaf_4.*C));
    d=(F_b.*D4^-1)         % amplitude(d) for mass4 from the system
    
   for d=(F_b.*D4^-1); 
   x= (abs(F_b./d)*1)
   omegaf_4=[0:1:80] 
   
   end

can u help me plot values from 80 different matrices as a function of omegaf_4?

[Jasneet Singh]

specifically for x w.r.t omega..

[Jan]

This is a new question. Please post it as a new thread.

Answer 2

[Walter Roberson]

n = 7;
MD = 2000*ones(1,n);
SD = -1000*ones(1,n-1);

K = diag(MD) + diag(SD,1) + diag(SD,-1)

K = 7×7

2000 -1000 0 0 0 0 0 -1000 2000 -1000 0 0 0 0 0 -1000 2000 -1000 0 0 0 0 0 -1000 2000 -1000 0 0 0 0 0 -1000 2000 -1000 0 0 0 0 0 -1000 2000 -1000 0 0 0 0 0 -1000 2000

Or:

n = 7;
K = zeros(n,7);
K(1:n+1:end) = 2000;
K(2:n+1:end) = -1000;
K(n+1:n+1:end) = -1000;
K

K = 7×7

2000 -1000 0 0 0 0 0 -1000 2000 -1000 0 0 0 0 0 -1000 2000 -1000 0 0 0 0 0 -1000 2000 -1000 0 0 0 0 0 -1000 2000 -1000 0 0 0 0 0 -1000 2000 -1000 0 0 0 0 0 -1000 2000