% 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