A = 2 -3 5 1 4 0 -1 3 2 1 2 3 First we check for existing orthogonality ans = -3 ans = 11 ans = -3 Pick one to start v1 = 2 1 -1 1 Then, remove the part of the second column that is parallel to v1 v2 = -2.1429 4.4286 2.5714 2.4286 Now, remove the parts of the third column that are parallel to v1 and v2 v3 = 1.9572 -1.7782 3.4514 1.3152 And check for orthogonality of all three vectors ans = 0 ans = 6.6613e-16 ans = 4.4409e-16 Finally, let's make all the vectors 'unit vectors' v1 = 0.75593 0.37796 -0.37796 0.37796 v2 = -0.35365 0.73088 0.42438 0.40081 v3 = 0.43086 -0.39146 0.75979 0.28953 And create a matrix of these three normalized and orthogonalized vectors that came from A Q = 0.75593 -0.35365 0.43086 0.37796 0.73088 -0.39146 -0.37796 0.42438 0.75979 0.37796 0.40081 0.28953 Notice that after this process, the 'A = QR' factorization can be obtained by Q'*A = R R = 2.64575 -1.13389 4.15761 0.00000 6.05923 0.28292 0.00000 -0.00000 4.54249 And we can check our factorization ans = 0.0000e+00 0.0000e+00 0.0000e+00 0.0000e+00 -1.7764e-15 4.4409e-16 -1.1102e-16 -4.4409e-16 -8.8818e-16 0.0000e+00 -4.4409e-16 0.0000e+00 Let's solve a system using Least Squares b = 4 -2 -3 1 mat = 2 -3 5 4 1 4 0 -2 -1 3 2 -3 1 2 3 1 ans = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ans = 1.00000 0.00000 0.00000 0.98246 0.00000 1.00000 0.00000 -0.62399 0.00000 0.00000 1.00000 0.11371 x = 0.98246 -0.62399 0.11371 r = -0.40543 -0.48652 -0.37300 0.92438 ans = 1.1810