How to invert a matrix that has variables?

Hello everybody, I'm trying to invert a matrix in which every entry depends on the variable h. If I experiment in a general case and do:

syms h
A=[1*h (2-h)*3;(2-h)*3 0]
B=inv(A)

I am able to recieve the inverse matrix of A because it is a square symmetric matrix. However, in my specific case I have:

 w=ones(2,1);
p=5;
x=w.^(1/p);
S=rand(1,2);
syms h
alpha=x+h*S';
w_alpha=alpha.^p;
K_total=sym(zeros(8));
C_1=rand(3);
C_1=(C_1+C_1')/2;
C_2=rand(3);
C_2=(C_2+C_2')/2;
C=w_alpha(1)*C_1+w_alpha(2)*C_2;
B=rand(3,8);
g_L_ele=[1 2 5 6 7 8 3 4];
K_total(g_L_ele,g_L_ele)=...
    K_total(g_L_ele,g_L_ele)+...
    B'*C*B;
K=vpa(K_total,2);
K=(K+K')/2;
inv(K)

Even though it's still a square symmetric matrix, that only depends on h, I can't get the inverse. Can anybody help me please? I have a final project due to tomorrow and this is the only part I'm missing in order to complete it.

[John D'Errico]

K_total is undefined, although you have defined a variable named K_total_h. So of course your code must fail.

Note that reducing the matrix to 2 digits is pure insanity when trying to compute an inverse. Could that also be part of your problem? How can I know?

[Ronaldinho Bato]

Oh my bad. K_total_h was supposed to be K_total in the code I'm sorry, already edited. However, it was just a typo passing it from my code to the website, it still doesn't work... I used only 2 digits precisely because it wasn't working with more and I thought it would help but no luck...

Answer 1

[Walter Roberson]

rank(C) is 3, so rank(B'*C*B) will be 3, so inv() can only work the vpa(K_total,2) accidentally introduces creates rank 8 out of the rank 3 matrix.

[Ronaldinho Bato]

I see... So is there any way to compute the inverse of the Matrix considering that I have a variable there or none at all? Because the same code but with a scalar instead of "h" gives a valid answer

[Walter Roberson]

You only get a result by accident due to floating point round off. Look at:

w = sym(ones(2,1));
p = sym(5);
x = w.^(1/p);
S = sym(rand(1,2));
%syms h real
h = sym(rand);
alpha = x+h*S';
w_alpha = alpha.^p;
K_total = sym(zeros(8));
C_1 = sym(rand(3));
C_1 = (C_1+C_1')/2;
C_2 = sym(rand(3));
C_2 = (C_2+C_2')/2;
C = w_alpha(1)*C_1 + w_alpha(2)*C_2;
B = sym(rand(3,8));
g_L_ele = [1 2 5 6 7 8 3 4];
K_total(g_L_ele,g_L_ele) = ...
    K_total(g_L_ele,g_L_ele) + ...
    B'*C*B;
K = double(K_total);
K=(K+K')/2;
rank_K = rank(K)
invK = inv(K)
rank_invK = rank(invK)

in this could, which runs with higher precision until just before doing the inv, you can see that rank(K) is 3 even though a scalar value was used for h. From this we can conclude that if K is rank 8 in your code, it is only due to loss of precision at some earlier step.