# 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.
1) Transposed matrix
2) Inverse matrix
##### Solution
(a) $(\bold{AB})^t = \bold{B}^t \bold{A}^t$$(\bold{AB})^t = \bold{B}^t \bold{A}^t$

Let $\bold{C} = \bold{AB}$$\bold{C} = \bold{AB}$, then $C_{ij} = [AB]_{ij} = \sum_k A_{ik}B_{kj}$$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}$$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$$(\bold{AB})^t = \bold{B}^t \bold{A}^t$
(b) $(\bold{AB})^{-1} = \bold{B}^{-1} \bold{A}^{-1}$$(\bold{AB})^{-1} = \bold{B}^{-1} \bold{A}^{-1}$

i) For orthogonal matrices, $\bold{\lambda}^t \bold{\lambda}^{-1}$$\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}$$(\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}$$(\bold{AB})^{-1} = \bold{B}^{-1} \bold{A}^{-1}$