Grassmann Algebra

TheExteriorProduct.nb 33
xi � C1 � � � Ci�1 � C0 � Ci�1 � � � Cn
�������������������������������� ��������������������������������
C1 � C2 � � � Cn
All the well-known properties of solutions to systems of linear equations proceed directly from
the properties of the exterior products of the Ci .
� Example solution
Consider the following system of 3 equations in 4 unknowns.
w � 2�x � 3�y � 4�z � 2
2�w � 7�y � 5�z � 9
w � x � y � z � 8
To solve this system, first declare the unknowns w, x, y and z as scalars, and then declare a
basis of dimension at least equal to the number of equations. In this example, we declare a 4space
so that we can add another equation later.
Next define:
DeclareExtraScalars��w, x, y, z��
�������
2.33
�a, b, c, d, e, f, g, h, w, x, y, z, �, �_ � _� ?InnerProductQ, _� 0
DeclareBasis�4�
�e1, e2, e3, e4�
Cw � e1 � 2�e2 � e3;
Cx ��2�e1 � e3;
Cy � 3�e1 � 7�e2 � e3;
Cz � 4�e1 � 5�e2 � e3;
C0 � 2�e1 � 9�e2 � 8�e3;
The system equation then becomes:
wCw � xCx � yCy � zCz � C0
Suppose we wish to eliminate w and z thus giving a relationship between x and y. To
accomplish this we multiply the system equation through by Cw � Cz .
�w Cw � xCx � yCy � zCz��Cw � Cz � C0 � Cw � Cz
Or, since the terms involving w and z will obviously be eliminated by their product with
Cw � Cz , we have more simply:
2001 4 5
�x Cx � yCy��Cw � Cz � C0 � Cw � Cz