Thornton & Marion, Classical Dynamics, 5th Edition

Chapter 1. Matrices, Vectors, and Vector Calculus

Problem 04. Transposed and Inverse Matrix

• prove the properties of the transposed and inverse matrix.

1. Transposed matrix
2. Inverse matrix

Solution

(a) $(\bold{AB})^t = \bold{B}^t \bold{A}^t$

Let $\bold{C} = \bold{AB}$, then $C_{ij} = [AB]_{ij} = \sum_k A_{ik}B_{kj}$

By definition of transpose, $C^t_{ij} = C_{ji} = [AB]_{ji} = \sum_k A_{jk} B_{ki} = \sum_k B_{ki} A_{jk} = \sum_k (B^t)_{ik} (A^t)_{kj} = [B^t A^t]_{ij}$

Therefore, $(\bold{AB})^t = \bold{B}^t \bold{A}^t$

(b) $(\bold{AB})^{-1} = \bold{B}^{-1} \bold{A}^{-1}$

i) For orthogonal matrices, $\bold{\lambda}^t \bold{\lambda}^{-1}$, so we can use the result of (a).

ii) For non-orthogonal matrices (i.e. In general), $(\bold{AB}) (\bold{AB})^{-1} = \bold{AB} (\bold{B}^{-1} \bold{A}^{-1}) = \bold{A} \bold{I} \bold{A}^{-1} = \bold{A} \bold{A}^{-1} = \bold{I}$

Therefore, $(\bold{AB})^{-1} = \bold{B}^{-1} \bold{A}^{-1}$