Think of that as 9 separate equations:

4a+ 4y+ zc= 1

a+ 4y= 1

9a+ b+ 4c= 1

a+ 4b+ 4c= 1

9a+ 9y+ 4c= 1

4a+ 4z= 1

a+ 4b+ 4c= 1

a+ b+ c= 1

9a+ 4b+ 4c= 1

By setting x, y, and z equal to each or those 9 sets of values in the equation .

Now write that as a matrix equation:

or AX= B.

A maps into a 3 dimensional subspace of . You can solve that equation if and only if B happens to lie in that subspace. If it does not, you can try to find X such that AX, while still in the subspace, is as close to B as possible (the "best fit"). That will be true when the difference, B- AX, is perpendicular to Ax. In fact, it will be orthogonal to the entire subspace. In terms of the inner product, <Au, B-AX>= 0 where u is any vector in . Let A' be the adjoint of A.

(For A a linear transformation from inner product space U to inner product space V, the "adjoint", A', is defined as the linear transformation from V back to U such that where the subscript indicates the space in which the inner product is taken. For a matrix with real number entries the adjoint is just the transpose.)

Then <Au, B- AX>= <u, A'(B- AX)>= 0. The difference now is that since u could beanyvector in , we must have A'(B- AX)= A'B- A'AX= 0 or A'AX= A'B. While A, being a "9 by 3" matrix is not invertible, A'A is a 3 by 3 matrix and may be invertible. If it is then is the "best fit". This is essentially a "least squares" best fit.