Let \(\mathbf{x}\) be a random vector and \(\boldsymbol{V}\) its variance matrix. Show that \(\mathbf{x}^{\top} \boldsymbol{V}^{-1} \mathbf{x}\) is invariant under linear transformation. More precisely, let \(\boldsymbol{A}\) be some nonsingular square matrix, \(\mathbf{x}^{\prime}=\boldsymbol{A} \mathbf{x}\), and \(\boldsymbol{V}^{\prime}\) the variance of \(\mathbf{x}^{\prime}\). Then, \(\mathbf{x}^{\prime \top}\left(\boldsymbol{V}^{\prime}\right)^{-1} \mathbf{x}^{\prime}=\mathbf{x}^{\top} \boldsymbol{V}^{-1} \mathbf{x}\).