- #1
hholzer
- 37
- 0
I am reading through David Widder's Advanced Calculus and he abbreviates a determinant
as:
[tex]
\left( \begin{array}{cccc}
r_{1} \ s_{1} \ t_{1}\\
r_{2} \ s_{2} \ t_{2}\\
r_{3} \ s_{3} \ t_{3}\\
\end{array} \right)
[/tex]
And refers to it by (rst). He then states that expanding by the minors of a given
column, we have:
[tex]
(rst) = r \cdot (s \times t) = s \cdot (t \times r) = t \cdot (r \times s)
[/tex]
Now, I worked it out by looking at the cofactors of r_1, r_2, and r_3 which are the components of the vector (s x t) and confirmed it holds. But how can I see
this property without having to do that? Is there another way to see this say algebraically or geometrically? Some more intuitive way, perhaps?
as:
[tex]
\left( \begin{array}{cccc}
r_{1} \ s_{1} \ t_{1}\\
r_{2} \ s_{2} \ t_{2}\\
r_{3} \ s_{3} \ t_{3}\\
\end{array} \right)
[/tex]
And refers to it by (rst). He then states that expanding by the minors of a given
column, we have:
[tex]
(rst) = r \cdot (s \times t) = s \cdot (t \times r) = t \cdot (r \times s)
[/tex]
Now, I worked it out by looking at the cofactors of r_1, r_2, and r_3 which are the components of the vector (s x t) and confirmed it holds. But how can I see
this property without having to do that? Is there another way to see this say algebraically or geometrically? Some more intuitive way, perhaps?