Matlab, operator A\B

 What is the result of the operation A\B, where A(1, m) and B (1, m)?

In the manual it is written:

A\B returns a least-squares solution to the system of equations A*x= B.

So it means x = inv (A'*A)*A'*B? However, the matrix A'*A is singular...

Let us suppose: 

 

A=[1 2 3]
B=[6 7 6]
A\B

0         0         0
0         0         0
2.0000    2.3333    2.0000
If ve use MLS:

C = inv (A'*A)   singular matrix
C = pinv(A'*A)

0.0051    0.0102    0.0153
0.0102    0.0204    0.0306
0.0153    0.0306    0.0459

D= C*A'*B

0.4286    0.5000    0.4286
0.8571    1.0000    0.8571
1.2857    1.5000    1.2857 

So results A\B and inv (A'*A)*A'*B are different...  

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Answers: 
My MATLAB (R2010b) says quite a lot about what A\B does: 
mldivide(A,B) and the equivalent A\B perform matrix left division (back slash). A and B must be matrices that have the same number of rows, unless A is a scalar, in which case A\B performs element-wise division — that is, A\B = A.\B.
If A is a square matrix, A\B is roughly the same as inv(A)*B, except it is computed in a different way. If A is an n-by-n matrix and B is a column vector with n elements, or a matrix with several such columns, then X = A\B is the solution to the equation AX = B. A warning message is displayed if A is badly scaled or nearly singular.
If A is an m-by-n matrix with m ~= n and B is a column vector with m components, or a matrix with several such columns, then X = A\B is the solution in the least squares sense to 
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