Hello all,
I am trying to invert a 5×5 matrix of mixed symbolics and numbers. All columns are linearly independent, concluded by inspection. However rank() returns 4 and the inverse of the matrix returns a 5×5 matrix where all elements are Inf.
Is there a better way to invert this matrix? It is not singular, as an inverse is returned, but it seems to be of indeterminate form.
Code:
% define coefficients
mu1 = sym('mu1');mu2 = sym('mu2');mu3 = sym('mu3');gam1 = sym('gam1');gam2 = sym('gam2');gam3 = sym('gam3');L = sym('L');% define matrices
B = [1 -1 -1 0 0 gam1/mu1 gam2/mu2 -gam2/mu2 0 0 0 exp(-gam2*L) exp(gam2*L) -exp(-gam3*L) -exp(gam3*L) 0 (-gam2/mu2)*exp(-gam2*L) (gam2/mu2)*exp(gam2*L) (gam3/mu3)*exp(-gam3*L) (-gam3/mu3)*exp(gam3*L) 0 0 0 0 0]rank(B)inv(B)
Best Answer