We can rewrite $$ \frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right), $$ as $$ \hat{A}^T\hat{b} = \hat{A}^T\hat{A}\hat{\beta}, $$ and if the matrix \( \hat{A}^T\hat{A} \) is invertible we have the solution $$ \hat{\beta} =\left(\hat{A}^T\hat{A}\right)^{-1}\hat{A}^T\hat{b}. $$