Thornton & Marion (Fifth Edition), Chapter 01, Problem 04 Solution

Thornton & Marion, Classical Dynamics, 5th Edition

Chapter 1. Matrices, Vectors, and Vector Calculus

Problem 04. Transposed and Inverse Matrix

The problem asks you to

  • prove the properties of the transposed and inverse matrix.

You should know about

  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}

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