Ax = b

Where x and b are column vectors, and A has the same number of rows as b, and x has less rows than both.

The least-squares method could be used here to obtain the best possible approximative solution. Let's call this solution "c".

Now, suppose I multiply some row of the equation system with a constant k. Let's say this row is the second row. In that case, I must multiply the 2nd row of A with k, as well as the 2nd row of b. This yields a new equation system, let's write it as:

Bx = d

If I use the method of least squares on the second system, I get a new solution that is different from c. Why is the solution different? Since I performed an elementary row operation on the first system to obtain the second system, shouldn't the two systems be equivalent, and therefore have the same least-squares solution?

When I did the same thing with a consistent system, I got the same solution for both systems.