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Mathematics 15 Online
OpenStudy (anonymous):

consider the matrix

OpenStudy (anonymous):

(a) Well, AA = AA so our first solution is B=A and AI=IA so the second is C=I (b) Now, combining these, AA+AI = AA+IA. We can factor out A (ensuring that we factor on the correct side..) A(A+I) = (A+I)A So we see that A+I is another solution. I'm not going to go into the full induction process but it's quite easy to show that A(A+A+I) = (A+A+I)A and carrying on.. A(kA+I) = (kA+I), k≥0. I'll admit that I lack the proof to demonstrate that these are the only solutions..

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