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**Ackbeet** You've correctly computed $\displaystyle A^{T}.$ Now form both products $\displaystyle AA^{T}$ and $\displaystyle A^{T}A.$ For the $\displaystyle AA^{T}$ product, note that $\displaystyle (I-B)$ and $\displaystyle (I+B)$ commute (just multiply it out both ways, and you'll see why). That will get you the result you want for that product. As for the $\displaystyle A^{T}A$ version, which you also should prove, that takes a bit more doing, because you really want elements with the inverses to commute also. Thinking...