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Unformatted text preview: % First, we can rewrite v = vP by v(P-I) = 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 = [P-I 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...
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This note was uploaded on 10/20/2010 for the course IEOR 4106 taught by Professor Whitt during the Spring '08 term at Columbia.
- Spring '08
- Operations Research