Page:Grundgleichungen (Minkowski).djvu/28

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then by AB, the product of the matrices A and B, will be denoted the matrix

C=\left|\begin{array}{ccc}
c_{11}, & \dots & c_{1r}\\
\vdots &  & \vdots\\
c_{p1}, & \dots & c_{pr}\end{array}\right|

where

c_{hk}=a_{h1}b_{1k}+a_{h2}b_{2k}+\dots+a_{hq}b_{qk}\quad\left({h=1,2,\dots p\atop k=1,2,\dots r}\right)

these elements being formed by combination of the horizontal rows of A with the vertical columns of B. For such a point, the associative law (AB)S = A(BS) holds, where S is a third matrix which has got as many horizontal rows as B (or AB) has got vertical columns.

For the transposed matrix of  C = AB, we have \bar{C}=\bar{B}\bar{A}.

3°. We shall have principally to deal with matrices with at most four vertical columns and for horizontal rows.

As a unit matrix (in equations they will be known for the sake of shortness as the matrix 1) will be denoted the following matrix (4 ✕ 4 series) with the elements.

(34) \left|\begin{array}{cccc}
e_{11}, & e_{12}, & e_{13}, & e_{14}\\
e_{21}, & e_{22}, & e_{23}, & e_{24}\\
e_{31}, & e_{32}, & e_{33}, & e_{34}\\
e_{41}, & e_{42}, & e_{43}, & e_{44}\end{array}\right| =\left|\begin{array}{cccc}
1, & 0, & 0, & 0\\
0, & 1, & 0, & 0\\
0, & 0, & 1, & 0\\
0, & 0, & 0, & 1\end{array}\right|

For a 4✕4 series-matrix, Det A shall denote the determinant formed of the 4✕4 elements of the matrix. If Det A \ne 0, then corresponding to A there is a reciprocal matrix, which we may denote by A^{-1} so that A^{-1} A = 1

A matrix

f=\left|\begin{array}{cccc}
0, & f_{12}, & f_{13}, & f_{14}\\
f_{21}, & 0, & f_{23}, & f_{24}\\
f_{31}, & f_{32}, & 0, & f_{34}\\
f_{41}, & f_{42}, & f_{43}, & 0\end{array}\right|,