% MATLAB Recitation Demo for Monday, September 15. % File: rdemo2a % % Example 1: % Recall definition for reduced row echelon form. % In MATLAB, write Ax = b as an augmented matrix. % Use MATLAB commands to convert [A b] into [R d]. % % Obtain solution x by "inspection" -- d is what % linear combination of the columns of R ? % Or, if you prefer, visualize backsubstitution % for Rx = d. % % Example 2: % Same b but change (3,3) entry in A. % >> diary rdemo2 >> b = [9; 29; 33] b = 9 29 33 >> A = [1 0 3; 3 2 9; 1 4 8] A = 1 0 3 3 2 9 1 4 8 >> Z = [A b] Z = 1 0 3 9 3 2 9 29 1 4 8 33 >> Z(2,:) = Z(2,:) - (3) * Z(1,:) Z = 1 0 3 9 0 2 0 2 1 4 8 33 >> Z(3,:) = Z(3,:) - (1) * Z(1,:) Z = 1 0 3 9 0 2 0 2 0 4 5 24 >> Z(3,:) = Z(3,:) - (2) * Z(2,:) Z = 1 0 3 9 0 2 0 2 0 0 5 20 >> Z(3,:) = (1/5) * Z(3,:) Z = 1 0 3 9 0 2 0 2 0 0 1 4 >> Z(2,:) = (1/2) * Z(2,:) Z = 1 0 3 9 0 1 0 1 0 0 1 4 >> Z(1,:) = Z(1,:) - (3) * Z(3,:) Z = 1 0 0 -3 0 1 0 1 0 0 1 4 % Z is now in reduced row echelon form; see definition. % d is the last column of Z. % R is the first 3 columns of Z. % We can solve Rx = d by inspection: d is obviously a % linear combination of columns of R. % Let's verify! >> x = [-3; 1; 4] x = -3 1 4 >> b b = 9 29 33 >> A * x ans = 9 29 33 %%%%%%%%%%%%%%%%% % Example 2 %%%%%%%%%%%%%%%%% % Let's change the (3,3) entry of A to 3. >> A(3,3) = 3 A = 1 0 3 3 2 9 1 4 3 >> Z = [A b] Z = 1 0 3 9 3 2 9 29 1 4 3 33 >> Z = ref(Z) Z = 1 0 3 0 0 1 0 0 0 0 0 1 % d is the last column of Z. % R is the first 3 columns of Z. % BAD news: Rx = d is not solvable because % d cannot be a linear combination of columns of R. % (Any linear combination of columns of R gives a vector % whose last component is zero.) % Or, note that the last equation of Rx = d is not % solvable because 0 * each "unknown" cannot add up % to 1. >> diary off