By trial and error find real non-zero 2 by 2 matrices such that:

1. A^2 = -I

2. BC = 0

3. DE = -ED (not allowing DE = 0)

== Solutions ==
Observe:
      |a b|^2   |a^2+bc ab+bd |        |-1  0|
A^2 = |   |   = |             | = -I = |     |
      |c d|     |ac+cd  bc+d^2|        |0  -1|
which means bc must be negative.  Trying the simplest case by setting a and d to zeros:
    |0  1|    
A = |    |
    |-1 0|
which turns out to work.  This means -A is also a solution.  Do you see why ?

We can interpret BC=0 as finding C such that the linear combination of the columns of B is zero.
Try a simple case when B is all ones:
     |1 1|| 1  1|
BC = |   ||     | = 0
     |1 1||-1 -1|
which is one solution.

This is the challenging one among the three.  In details, it means:
|a b||a' b'|    |a' b'||a b|
|   ||     | = -|     ||   |
|c d||c' d'|    |c' d'||c d|
This means 4 conditions must be satisfied:
2aa'      + bc' + b'c = 0
2dd'      + bc' + b'c = 0
ab' + ab' + bd' + b'd = 0
ac' + a'c + cd' + c'd = 0
From the first 2 equations, it also means aa' must equal dd'.
For example, if a=a'=1, dd' must be 1.  Let's try d=d'=-1:
|1   ||1   |
|  -1||  -1|
Now 2aa' = 2, (bc' + b'c) must equal -2 for the first equation to hold.
Try the simple case when bc' = -2 and b'c = 0, which implies either b' or c must be zero.
Let's try c = 0:
|1  1||1  b'|
|0 -1||-2 -1|
whatever b' value is, this turns out to work!  We can check the result using Octave:
D = [1 1; 0 -1]
E = [1 rand(); -2 -1]
Both:
D*E != 0
D*E == -E*D
would always result in all ones :-)

# posted by rot13(Unafba Pune) @ 3:45 PM