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

Chapter 01. Matrices, Vectors, and Vector Calculus

Problem 04. Transposed and Inverse Matrix

The problem asks you to
prove the properties of 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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