Unformatted text preview: % 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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- Spring '11
- Linear Algebra, stationary probability vector, chain transition matrix