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