Worksheet21: Dickson 110
Current Work (editable)
Similarly the value of the determinant (7) of order 4 may be found by expansion according to the elements of the fourth column: -d_{1} |a_{2} b_{2} c_{2} a_{3} b_{3} c_{3} a_{4} b_{4} c_{4}| + d_{2} |a_{1} b_{1} c_{1} a_{3} b_{3} c_{3} a_{4} b_{4} c_{4}| - d_{3} |a_{1} b_{1} c_{1} a_{2} b_{2} c_{2} a_{4} b_{4} c_{4}| + d_{4} |a_{1} b_{1} c_{1} a_{2} b_{2} c_{2} a_{3} b_{3} c_{3}|. We shall now prove that any determinant D of order n may be expanded according to the elements of any row or any column. Let E_{ij} denote the minor of e_{ij} in D, given by (8), so that E_{ij} is obtained by erasing the ith row and jth column of D. (i) We first prove that (10) D = e_{11} E_{11} - e_{21} E_{21} + e_{31} E_{31} - ... + (-1)^{n-1} e_{n1} E_{n1}, so that D may be expanded according to the elements of its first column. By (9) the terms of D having the factor e_{11} are of the form (-1)^{i} e_{11} e_{i_{2}2} ... e_{i_{n}n}, where 1, i_{2}, ..., i_{n} is an arrangement of 1, 2, ..., n, obtained from the latter by i interchanges, so that i_{2}, ..., i_{n} is an arrangement of 2, ..., n, derived from the latter by i interchanges. After removing from each term the common factor e_{11} and adding the quotients, we obtain a sum which, by definition, is the value of the determinant E_{11} of order n - 1. Hence the terms of D having the factor e_{11} may all be combined into e_{11} E_{11}, which is the first part of (10). We next prove that the terms of D having the factor e_{21} may be combined into -e_{21} E_{21}, which is the second part of (10). For, if \Delta be the determinant obtained from D by interchanging its first and second rows, the result just proved shows that the terms of \Delta having the factor e_{21} may be combined into the product of e_{21} by the minor |e_{12} e_{13} ... e_{1n} e_{32} e_{33} ... e_{3n} ......... e_{n2} e_{n3} ... e_{nn}| of e_{21} in \Delta. Now this minor is identical with the minor E_{21} of e_{21} in D. But \Delta = -D (§87). Hence the terms of D having the factor e_{21} may be
Answer Key (non-editable)
\begin{Remark}%[** Continued?] Similarly the value of the determinant \Eq{(7)} of order $4$ may be found by expansion according to the elements of the fourth column: \[ -d_{1} \left|\begin{array}{ccc} a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \\ a_{4} & b_{4} & c_{4} \end{array}\right| % + d_{2} \left|\begin{array}{ccc} a_{1} & b_{1} & c_{1} \\ a_{3} & b_{3} & c_{3} \\ a_{4} & b_{4} & c_{4} \end{array}\right| % - d_{3} \left|\begin{array}{ccc} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{4} & b_{4} & c_{4} \end{array}\right| % + d_{4} \left|\begin{array}{ccc} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|. \] \end{Remark} We shall now prove that \begin{Thm}any determinant $D$ of order $n$ may be expanded according to the elements of any row or any column\end{Thm}. Let $E_{ij}$ denote the minor of $e_{ij}$ in $D$, given by \Eq{(8)}, so that $E_{ij}$ is obtained by erasing the $i$th row and $j$th column of $D$. %[** Attn item label] (\textit{i}) We first prove that \[ \Tag{(10)} D = e_{11} E_{11} - e_{21} E_{21} + e_{31} E_{31} - \dots + (-1)^{n-1} e_{n1} E_{n1}, \] so that $D$ may be expanded according to the elements of its first column. By \Eq{(9)} the terms of $D$ having the factor $e_{11}$ are of the form \[ (-1)^{i} e_{11} e_{{i_{2}}2} \dots e_{i_{n}n}, \] where $1$, $i_{2}$, \dots, $i_{n}$ is an arrangement of $1$, $2$, \dots, $n$, obtained from the latter by $i$ interchanges, so that $i_{2}$, \dots, $i_{n}$ is an arrangement of $2$, \dots, $n$, derived from the latter by $i$ interchanges. After removing from each term the common factor $e_{11}$ and adding the quotients, we obtain a sum which, by definition, is the value of the determinant $E_{11}$ of order $n - 1$. Hence the terms of $D$ having the factor $e_{11}$ may all be combined into $e_{11} E_{11}$, which is the first part of \Eq{(10)}. We next prove that the terms of $D$ having the factor $e_{21}$ may be combined into $-e_{21} E_{21}$, which is the second part of \Eq{(10)}. For, if $\Delta$ be the determinant obtained from $D$ by interchanging its first and second rows, the result just proved shows that the terms of $\Delta$ having the factor $e_{21}$ may be combined into the product of $e_{21}$ by the minor \[ \left|\begin{array}{cccc} e_{12} & e_{13} & \dots & e_{1n} \\ e_{32} & e_{33} & \dots & e_{3n} \\ \hdotsfor{4} \\ e_{n2} & e_{n3} & \dots & e_{nn} \end{array}\right| \] of $e_{21}$ in $\Delta$. Now this minor is identical with the minor $E_{21}$ of $e_{21}$ in $D$. But $\Delta = -D$ (§87). Hence the terms of $D$ having the factor $e_{21}$ may be
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