Chapter 2 - Matrices and Systems of Linear Equations - 2.6 The Inverse of a Square Matrix - Problems - Page 178: 31

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Since the multiplication in matrices are associative, we have: $B^{-1}(A^{-1}A)B=I$ With $A^{-1}A=AA^{-1}=I$, then $B^{-1}(A^{-1}A)B=B^{-1}IB=B^{-1}B=I$ Hence, $(AB)^{-1}=B^{-1}A^{-1}$ Let $B=A^{-1}$ According to property $(AB)^T=B^TA^T$, we obtain: $(A^{-1})^TA^T=(AA^{-1})^T=I^T=I$ Consequently, $(A^{-1})^T=(A^T)^{-1}$