>> c1s6
Exercise number (1-14)? 4
A =
    0.3500   -0.3000   -0.3000   -0.2000         0
   -0.1000    0.9000   -0.1500   -0.1000         0
   -0.2500   -0.3500    0.8500   -0.3000         0
         0   -0.2500   -0.4000    0.6000         0
>> A(:,1:4)
ans =
    0.3500   -0.3000   -0.3000   -0.2000
   -0.1000    0.9000   -0.1500   -0.1000
   -0.2500   -0.3500    0.8500   -0.3000
         0   -0.2500   -0.4000    0.6000

>> % Taking the first four columns of A, and ignoring the last column of zeros

>> % We can use the above command to remove the last column of A:
>> A=A(:,1:4)                                                                 
A =
    0.3500   -0.3000   -0.3000   -0.2000
   -0.1000    0.9000   -0.1500   -0.1000
   -0.2500   -0.3500    0.8500   -0.3000
         0   -0.2500   -0.4000    0.6000
>> A=-A
A =
   -0.3500    0.3000    0.3000    0.2000
    0.1000   -0.9000    0.1500    0.1000
    0.2500    0.3500   -0.8500    0.3000
         0    0.2500    0.4000   -0.6000
>> % just change the sign of every entry in A; Now we get the coefficient matrix in the Lecture notes

>> Arref = rref(A)
Arref =
    1.0000         0         0   -2.0279
         0    1.0000         0   -0.5311
         0         0    1.0000   -1.1681
         0         0         0         0

>> % Let us get p_A as the free variable - we need to have the column
>> % for p_A sitting last, i.e., 4th in the A matrix

>> A(:,[2 3 4 1])
ans =
    0.3000    0.3000    0.2000   -0.3500
   -0.9000    0.1500    0.1000    0.1000
    0.3500   -0.8500    0.3000    0.2500
    0.2500    0.4000   -0.6000         0
>> A
A =
   -0.3500    0.3000    0.3000    0.2000
    0.1000   -0.9000    0.1500    0.1000
    0.2500    0.3500   -0.8500    0.3000
         0    0.2500    0.4000   -0.6000
>> B=A(:,[2 3 4 1])
B =
    0.3000    0.3000    0.2000   -0.3500
   -0.9000    0.1500    0.1000    0.1000
    0.3500   -0.8500    0.3000    0.2500
    0.2500    0.4000   -0.6000         0
>> Brref = rref(B)
Brref =
    1.0000         0         0   -0.2619
         0    1.0000         0   -0.5760
         0         0    1.0000   -0.4931
         0         0         0         0
>> Arref
Arref =
    1.0000         0         0   -2.0279
         0    1.0000         0   -0.5311
         0         0    1.0000   -1.1681
         0         0         0         0
>> 1/2.03
ans =
    0.4926
>> 1/2.0279 
ans =
    0.4931
>> 0.53/2.03
ans =
    0.2611
>> % Can do a row swap ERO more directly using this method
>> B
B =
    0.3000    0.3000    0.2000   -0.3500
   -0.9000    0.1500    0.1000    0.1000
    0.3500   -0.8500    0.3000    0.2500
    0.2500    0.4000   -0.6000         0
>> % ERO - swap R1 and R4 in B
>> B([4 2 3 1],:)
ans =
    0.2500    0.4000   -0.6000         0
   -0.9000    0.1500    0.1000    0.1000
    0.3500   -0.8500    0.3000    0.2500
    0.3000    0.3000    0.2000   -0.3500
>> B=B([4 2 3 1],:)
B =
    0.2500    0.4000   -0.6000         0
   -0.9000    0.1500    0.1000    0.1000
    0.3500   -0.8500    0.3000    0.2500
    0.3000    0.3000    0.2000   -0.3500

>> % saying 'help CMD' will print the help for any command CMD
>> help rref