52nd Putnam 1991

Problem A2

M and N are real unequal n x n matrices satisfying M3 = N3 and M2N = N2M. Can we choose M and N so that M2 + N2 is invertible?



Answer: No.


The only difficulty is that we are not told whether or not it is true. So one has to (1) try experimenting a little to find invertible M, N satisfying the conditions, and (2) try proving you cannot find such M, N. If you have bad luck you will try in that order!

(M2 + N2)M = M3 + N2M = N3 + M2N = (M2 + N2)N. But now if M2 + N2 was invertible, we could multiply by its inverse to get M = N, whereas we are told M and N are unequal.



52nd Putnam 1991

© John Scholes
21 Sep 1999