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

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′.