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

Thornton & Marion, Classical Dynamics of Particles and Systems, 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


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

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

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

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

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

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

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

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

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