Solving Systems of Equations Using MatricesDate: 04/02/99 at 09:55:14 From: Brian Subject: 2x2 idempotent matrices Dr. Math, My system of equations is: a^2 + bc = a ab + bd = b ac + cd = c bc + d^2 = d So far, I have basically been plugging in numbers to find the right combinations. I have found 8 matrices, all of which consist of 0's and 1's. I have also found a pattern where: [1-j 1-j] [j j ] is idempotent where j is any real number. Are these all of them? If not, should I be doing something else to try to figure them out? Thanks, Brian Date: 04/02/99 at 15:15:22 From: Doctor Rob Subject: Re: 2x2 idempotent matrices Thanks for writing to Ask Dr. Math! Write the equations in this form: a*(1-a) = b*c, b*(a+d-1) = 0, c*(a+d-1) = 0, d*(1-d) = b*c. Now there are several cases to consider. Case 1: a + d - 1 is nonzero. This implies that b = c = 0 from the second and third equations. Then from the first and last equations, either a = 0 or a = 1, and either d = 0 or d = 1. Since a + d - 1 is nonzero, you get either a = d = 0 or a = d = 1. This gives two matrices: (0 0) (1 0) (0 0) (0 1). Case 2: a + d - 1 = 0, so d = 1 - a. Subcase 2a: a = 0, so d = 1 and b*c = 0. This gives two classes of matrices: (0 0) (0 b) (c 1) (0 1) Subcase 2b: a = 1, so d = 0 and b*c = 0. This also gives two classes of matrices: (1 0) (1 b) (c 0) (0 0) Subcase 2c: a is neither 0 nor 1. This implies by the first equation that b cannot be zero, and c = a*(1-a)/b. This gives one final class of matrices: ( a b ) (a*(1-a)/b 1-a) The eight matrices are just the first six classes with the variables set equal to 0 or 1. Your pattern is just the last class with a = b = 1-j. An example of the last class is (5 -2) (10 -4) - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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