If \( \bar{A}^p = \frac{\partial \bar{x}^p}{\partial x^q} A^q \) prove that \( A^q = \frac{\partial x^q}{\partial \bar{x}^p} \bar{A}^p \).

Another Explanation (5): Given \( \bar{A}^p = \frac{\partial \bar{x}^p}{\partial x^q} A^q \), we can express this in matrix notation as \( \bar{\mathbf{A}} = \mathbf{J} \mathbf{A} \), where \( \mathbf{J} \) is the Jacobian matrix with elements \( J^p_q = \frac{\partial \bar{x}^p}{\partial x^q} \). Since the Jacobian matrix is invertible (assuming it represents a coordinate transformation), we can multiply both sides by the inverse Jacobian, \( \mathbf{J}^{-1} \), to obtain \( \mathbf{A} = \mathbf{J}^{-1} \bar{\mathbf{A}} \). The elements of the inverse Jacobian are given by \( (\mathbf{J}^{-1})^q_p = \frac{\partial x^q}{\partial \bar{x}^p} \). Therefore, \( A^q = \frac{\partial x^q}{\partial \bar{x}^p} \bar{A}^p \).