On Wed Nov 3 22:06:01 EDT 2010

	 [2    -3    0]
   M0 := [0     3    2] ,        TraceM0 = 12
	 [8     1    7]


	 [1    5    -1]
   C1 := [0    2     5]
	 [0    1     3]

> M1 := MI(C1) . M0 . C1 ;  TraceM1 := Trace(M1) ;

	 [218    1199    133]
   M1 := [-40    -221    -27] ,  TraceM1 = 12
	 [ 16      90     15]


	 [27     5     2]
   C2 := [ 5    29    17]
	 [ 2    17    10]

> M2 := MI(C2) . M0 . C2 ;  TraceM2 := Trace(M2) ;

	 [   6080      3063      1598]
   M2 := [-101085    -50994    -26609] ,  TraceM2 := 12
	 [ 170652     86096     44926]


## So each two of M0, M1, M2 are conjugate-equivalent.  And we saw
that each has trace=12.
  Now let's compute their characteristic-polynomials.  Firstly,

	    [1    0    0]
      I3 := [0    1    0]
	    [0    0    1]

>  charpoly0 := Det(M0 - x*I3);
                                                   2    3
                     charpoly0 := -10 - 39 x + 12 x  - x

>  charpoly1 := Det(M1 - x*I3);
                                                   2    3
                     charpoly1 := -10 - 39 x + 12 x  - x

>  charpoly2 := Det(M2 - x*I3);
                                                   2    3
                     charpoly2 := -10 - 39 x + 12 x  - x

## Indeed, the three char-polys are equal.
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