Unformatted text preview: % First, we can rewrite v = vP by v(PI) = z, where I is an identity matrix and z is a vector of zeros. % We then add a column of 1's to make a new equation % I = eye(n); %the identity matrix z = zeros(1,n); %a row of zeros w = ones(n,1); %a column of 1's A = [PI w]; % % The desired system of equations is vA = [z 1], where A is n by (n+1) % We solve it by writing v = [z 1]/A % v = [z 1]/A; % % Using transposes, we could also write v' = A'\[z 1]' % We could also use the matrix inverse applied to square matrices. % That approach is in the other program stationary.m...
View
Full Document
 Spring '08
 Whitt
 Linear Algebra, Operations Research, Matrices, Ring, Markov chain, Invertible matrix

Click to edit the document details