Date: 06/01/98 at 06:32:02 From: Andrea Quintiliani Subject: a problem in matrix algebra Hello, I've been trying to help my son with his school work, and have come upon a problem that I've not been able to solve. I'm a graduate in physics and have asked other colleagues, but nobody seems to come up with a correct solution (everybody got rusty since University!). I also searched your database with no success. The problem is this: Let A, B, C, X be matrices, for instance 2x2 matrices. Calculate X in the expression: A*X - X*B = C For example: [1 2] * X - X * [2 2] = [2 4] [2 1] [-1 0] [0 -1] If it were: A*X + B*X = C then (A + B) * X = C, and setting A + B = D, D!= inverse matrix of D, we have, D!*D*X = D!*C. Thus: I*X = X = D!*C where I = identity matrix But, given also the non-commutability of matrix products, I guess I cannot apply this grouping technique to the problem at hand. I am really curious about finding a solution to this problem and hope that somebody will find the time to help me. Thank you very much, Andrea Quintiliani
Date: 06/01/98 at 11:19:47 From: Doctor Anthony Subject: Re: a problem in matrix algebra Using your values in the equation A*X - X*B = C we get: [1 2]*[a b] - [a b]*[ 2 2] = [2 4] [2 1] [c d] [c d] [-1 0] [0 -1] [a+2c b+2d] - [2a-b 2a] = [2 4] [2a+c 2b+d] [2c-d 2c] [0 1] [-a+b+2c -2a+b+2d] = [2 4] [ 2a-c+d 2b-2c+d] [0 1] So we get four equations: -a + b + 2c + 0*d = 2 -2a + b + 0*c + 2*d = 4 2a + 0*b - c + d = 0 0*a + 2b - 2c + d = 1 And we obtain a = -2/5, b = 2/5, c = 3/5, d = 7/5; so the matrix X is: X = (1/5)[-2 2] [ 3 7] -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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