Page:Grundgleichungen (Minkowski).djvu/28

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

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