Why not add more variables to obtain an invertible matrix, in order to solve the SAT problem in polynomial time?

[Z.10.46]

 

Hello,

 

In short, we can transform the SAT problem into AX=1 withX a vector and 1 the vector of 1, but the determinant A is often zero, so A^-1 does not exist.

However, assuming that since I have the form of the A determinant, I could construct a polynomial-time algorithm where I add more variables to avoid having a zero determinant, I could then find 𝐴′^-1, and the solution would be in polynomial time to find X': X in X′.