If [tex] A = [a_{ij}]^{n\times n} [/tex] is invertible, show that [tex] (A^{2})^{-1} = (A^{-1})^{2} [/tex] and [tex] (A^{3})^{-1} = (A^{-1})^{3} [/tex](adsbygoogle = window.adsbygoogle || []).push({});

So basicaly we have a square matrix with elements [tex] a_{ij} [/tex]. This looks slightly familar to [tex] (A^{T})^{-1} = (A^{-1})^{T} [/tex]. Are [tex] A^{2} [/tex] and [tex] A^{3} [/tex] meant to be the elements of the matrix raised to those respective powers? Or does it mean that the matrix is [tex] 2\times 2 [/tex] or [tex] 3\times 3 [/tex]?

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# Homework Help: Invertible Matrices

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