**Problem**: (2003 IMC Day 2 - #1) Let $ A $ and $ B $ be $ n \times n $ real matrices such that $ AB+B+A = 0 $. Prove that $ AB = BA $.

**Solution**: Well, seeing the $ AB+B+A $, we think Simon's Favorite Factoring Trick. But, we have to remember that we're working with matrices, not numbers. So instead of $ (x+1)(y+1) = xy+x+y+1 $, we have

$ (A+I_n)(B+I_n) = AB + I_nB+AI_n+I_n^2 $,

where $ I_n $ is the $ n \times n $ identity matrix (the equivalent of $ 1 $ in the matrix world). Recall that for any square matrix $ M $ we have $ MI = IM = M $ (where $ I $ is the identity matrix of corresponding size), so

$ AB + I_nB+AI_n+I_n^2 = AB+B+A+I_n = 0+I_n = I_n $

from the given condition. But the equation

$ (A+I_n)(B+I_n) = I_n $

means that $ A+I_n $ and $ B+I_n $ are inverses of each other. Which consequently means we can multiply them in either order. So

$ (A+I_n)(B+I_n) = (B+I_n)(A+I_n) $

$ AB+B+A+I_n = BA+A+B+I_n \Rightarrow AB = BA $

as desired. QED.

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Comment: Just basic algebraic manipulations if you knew several key properties of matrices and recognized the factoring. Always good to keep these things handy!

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Practice Problem: (1999 IMC Day 1 - #1) Show that for all natural numbers $ n $ there exists a real $ n \times n $ matrix $ A $ such that $ A^3 = A+I $. Furthermore, prove that $ |A| > 0 $ for all such $ A $. [Reworded]

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