$$\det(\mathbf{A}) = \begin{bmatrix}1&0\end{bmatrix}\mathbf{A}^T\begin{bmatrix}0&1\\1&0\end{bmatrix}\mathbf{A}\begin{bmatrix}0\\1\end{bmatrix}.$$
Prove this and extend it to general $n \times n$ matrices. Can you go further than that?
Another old idea. Given a $2 \times 2$ matrix $\mathbf{A}$,
$$\det(\mathbf{A}) = \begin{bmatrix}1&0\end{bmatrix}\mathbf{A}^T\begin{bmatrix}0&1\\1&0\end{bmatrix}\mathbf{A}\begin{bmatrix}0\\1\end{bmatrix}.$$ Prove this and extend it to general $n \times n$ matrices. Can you go further than that?
0 Comments
Leave a Reply. 
SharingBesides sharing my own musings and insights on various topics, I recommend some books, events or other material. Archives
May 2016
Except where otherwise noted, site content created by Cheng Herng Yi is licensed under a Creative Commons AttributionNonCommercialShareAlike 4.0 International License. Categories
All
