Linear least-squares method and row multiplication of matrix

[Mesud1]

Suppose that I have an overdetermined equation system in matrix form:

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.

 

[DrDu]

The least squares method minimizes the sum of the deviations of the left hand from the right hand side. If you multiply one equation by c, this equation gets more weight in the sum and the optimal solution will be different. This doesn't happen if all equations can be fulfilled identically.

 

[Mesud1]

The least squares method minimizes the sum of the deviations of the left hand from the right hand side. If you multiply one equation by c, this equation gets more weight in the sum and the optimal solution will be different. This doesn't happen if all equations can be fulfilled identically.

Makes perfect sense, thank you.