I've been trying to prove that $D(B, C) = -D(C, B)$ by using the components of the matrix and I need help on one specific part. I'll put down what I got below and let me know if it is correct. If it is, please point me to a property of determinants that I may have missed because I am confused.
I started by defining B and C as such: $$ B=\begin{pmatrix}b_1\\b_2\end{pmatrix} \\ C=\begin{pmatrix}c_1\\c_2\end{pmatrix} $$ Then I moved $-D(C,B)$ over to get this: $D(B,C) + D(C,B) = 0$
However, I want to do it in matrix component form. I wrote it as such:
$ \det\begin{pmatrix} b_1 & c_1 \\ b_2 & c_2 \end{pmatrix} + \det\begin{pmatrix} c_1 & b_1 \\ c_2 & b_2 \end{pmatrix}=0 $
I am following the proof in the form D(B,C) and now here is my question.
Is the following true?
$ \det\begin{pmatrix} b_1 & c_1 \\ b_2 & c_2 \end{pmatrix} + \det\begin{pmatrix} c_1 & b_1 \\ c_2 & b_2 \end{pmatrix}=\det\begin{pmatrix} b_1+c_1 & b_1+c_1 \\ b_2+c_2 & b_2+c_2 \end{pmatrix} $
This is all I need help on, the rest I can do on my own. Thanks.
Partial Answer
The claim is not correct. $$\mathrm{det}(A) + \mathrm{det}(B) \ne \mathrm{det}(A + B)$$ There are other ways to see this but as a counter example let us take a 2-by-2 example $A=B=\mathrm{id}(=I)$ (the identity matrix). Clearly: $$ \mathrm{det}(I) + \mathrm{det}(I) = 2\mathrm{det}(I) = 2 $$ Wheras: $$ \mathrm{det}(I+I) = \mathrm{det}(2I) = 4 $$ The property you are looking at is called the skew symmetry of the determinant. To prove it you can do this a number of ways. You can do this by either using the expanded form of the determinant ( a bit complicted in my opinion ) or you can simply evaluate for an arbirary pair of vectors as you have started. A more geometric way to see the sign change is from the cross product. Given 2 vectors in 2D $a,b$ and the signed angle swept from $a$ to $b$ is $\theta(a,b)$: $$ D(a,b) = |a||b|\sin(\theta(a,b)) = | a \times b |$$ It follows then that switching $a$ and $b$ gives the negative angle in going from $b$ to $a$. Since the sine function is odd, the result correspondingly picks up a minus sign.