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### Solving Systems of Equations Using Matrices

```
Date: 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/
```
Associated Topics:
High School Linear Algebra

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