Tilia Posted May 21, 2009 Report Share Posted May 21, 2009 (edited) I have a test next week, and since my teacher isn't used to teaching SL, I hope someone here can help me more than she can.1. "It is known that AB=A and BA=A where A and B are not necessarily invertible. Prove that A^2=A." Why isn't B=I?2. "Write 5A^2-6A=2I in the form AB=I and hence find A^-1 in terms of A and IThanks in advance! Edited May 21, 2009 by Tilia Reply Link to post Share on other sites More sharing options...
Hedron123 Posted May 23, 2009 Report Share Posted May 23, 2009 I have a test next week, and since my teacher isn't used to teaching SL, I hope someone here can help me more than she can.1. "It is known that AB=A and BA=A where A and B are not necessarily invertible. Prove that A^2=A." Why isn't B=I?2. "Write 5A^2-6A=2I in the form AB=I and hence find A^-1 in terms of A and IThanks in advance!Okay, I'll try to solve this.(i) As AB = A and BA = B and you have to prove that A^2 = A.Then:(AB) x (BA) = ABA^2 B^2 = AB(AB)^2 = ABBut AB = A, hence.(A)^2 = A(ii) Multiplying a matrix by the identity matrix is the same as multiplying the matrix by 1, that is, you obtain the same matrix as a result. This would be true for:AB = AA x I = A.However, A is equal to A^2 not to A. Thus, B cannot be I.I don't know how to solve part 2. Are you sure this is SL maths? I have never seen an exercise of these characteristics for SL. Reply Link to post Share on other sites More sharing options...
Tilia Posted May 24, 2009 Author Report Share Posted May 24, 2009 Okay, I'll try to solve this.(i) As AB = A and BA = B and you have to prove that A^2 = A.Then:(AB) x (BA) = ABA^2 B^2 = ABWhy? I don't know how to solve part 2. Are you sure this is SL maths? I have never seen an exercise of these characteristics for SL.Well, it was in my SL maths book at least, and that's the only reference I have. Reply Link to post Share on other sites More sharing options...
sweetnsimple786 Posted May 24, 2009 Report Share Posted May 24, 2009 (edited) I have a test next week, and since my teacher isn't used to teaching SL, I hope someone here can help me more than she can. 1. "It is known that AB=A and BA=A where A and B are not necessarily invertible. Prove that A^2=A." Why isn't B=I? 2. "Write 5A^2-6A=2I in the form AB=I and hence find A^-1 in terms of A and I Thanks in advance! Okay, I'll try to solve this. (i) As AB = A and BA = B and you have to prove that A^2 = A. Then: (AB) x (BA) = AB A^2 B^2 = AB (AB)^2 = AB But AB = A, hence. (A)^2 = A (ii) Multiplying a matrix by the identity matrix is the same as multiplying the matrix by 1, that is, you obtain the same matrix as a result. This would be true for: AB = A A x I = A. However, A is equal to A^2 not to A. Thus, B cannot be I. I don't know how to solve part 2. Are you sure this is SL maths? I have never seen an exercise of these characteristics for SL. Wait... BA is equal to A, not B. So wouldn't that mean that... A=AB A=BA so AB=BA The only way that is possible (if B isn't the identity matrix) is if A is the zero matrix. When you square the zero matrix, you get the zero matrix so A^2=A Edited May 24, 2009 by sweetnsimple786 Reply Link to post Share on other sites More sharing options...
sweetnsimple786 Posted May 24, 2009 Report Share Posted May 24, 2009 Okay So I've been stewing over the second part for a lil bit and here's what I've got.. please tell me if it makes sense:AB=I1) recognize that B=A^-12) Substitute 5A^2 - 6A = 2AB3) Multiply each side by A^-1 5A - 6I = 2B4) Divide by 2 2.5A - 3I =B5) B=A^-1, so I think that's the answer 2.5A-3I=B Reply Link to post Share on other sites More sharing options...
Tilia Posted May 25, 2009 Author Report Share Posted May 25, 2009 The correct answer is A^-1=(5/3)A - 2ISuppose you're on the right track. Thanks Reply Link to post Share on other sites More sharing options...
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