- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
3.2 $\quad$ Motivation for the choice of axioms for a determinant function
$\quad$ In Volume 1, we proved that the scalar triple product of three vectors $A_1, A_2, A_3$ in $V_3$ can be expressed as the determinant of a matrix whose rows are the given vectors. Thus, we have \begin{align*} A_1 \times A_2 \cdot A_3 &= \det\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \end{align*} where $A_1 = (a_{11}, a_{12}, a_{13}), A_2 = (a_{21}, a_{22}, a_{23}),$ and $A_3 = (a_{31}, a_{32}, a_{33}.$)
$\quad$ If the rows are linearly independent, the scalar triple product is nonzero and its absolute value is the volume of the parallelepiped determined by the vectors $A_1, A_2,$ and $A_3.$ If the rows are dependent, the scalar triple product is zero and the parallelepiped degenerates into a plane figure of zero volume.
$\quad$ Some of the properties of the scalar triple product will serve as motivation for the choice of axioms for a determinant function in the higher-dimensional case. To state these properties in a form suitable for generalization, we consider the scalar triple product as a function of the three row vectors $A_1, A_2, A_3.$ We denote this function by $d;$ thus
\begin{align*}
d(A_1, A_2, A_3) &= A_1 \times A_2 \cdot A_3.
\end{align*}
We focus our attention on the following properties:
$\quad$ (a) $\quad$ Homogeneity in each row. $\quad$ For example, homogeneity in the first row states that
\begin{align*}
d(tA_1, A_2, A_3) &= t\,d(A_1, A_2, A_3) \qquad \text{for every scalar $t.$}
\end{align*}
$\quad$ (b) $\quad$ Additivity in each row $\quad$ For example, additivity in the second row states that
\begin{align*}
d(A_1, A_2 + C, A_3) &= d(A_1, A_2, A_3) + d(A_1, C, A_3)
\end{align*}
for every vector $C.$
$\quad$ (c) $\quad$ The scalar triple product is zero if two of the rows are equal.
$\quad$ (d) $\quad$ Normalization.
\begin{align*}
d(\mathbf{i, j, k}) &= 1 \quad \text{where} \quad \mathbf{i} = (1, 0, 0), \quad \mathbf{j} = (0, 1, 0), \quad \mathbf{k} = (0, 0, 1).
\end{align*}