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