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Exercise 2.7.2

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${(AB)}^{T} = [\begin{matrix} 1 & 2 \\ 3 & 7 \end{matrix}] = B^{T} A^{T} .$ This answer is different from $A^{T} B^{T}$ (except when $AB = BA$ and transposing gives $B^{T} A^{T} = A^{T} B^{T}$ ).

mathchakra

2021-12-20 21:25

Exercise 2.7.2

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