determinant

Printable View

Aug 19th 2007, 06:52 AM
needhelpsobad

1 Attachment(s)

determinant

ahhhh
Aug 19th 2007, 07:29 AM
topsquark

Quote:

Originally Posted by needhelpsobad

ahhhh

$\left ( \begin{matrix} a & b \\ c & d \end{matrix} \right ) \cdot \left ( \begin{matrix} e & f \\ g & h \end{matrix} \right ) = \left ( \begin{matrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{matrix} \right )$

$\text{det} \left ( \begin{matrix} a & b \\ c & d \end{matrix} \right ) = ad - bc$

-Dan
Aug 20th 2007, 05:14 AM
curvature

Quote:

Originally Posted by needhelpsobad

ahhhh

For (c), you can use the equality in (b) to show that AB is singular since det(AB)=0.

All times are GMT -8. The time now is 07:30 AM.

Copyright © 2005-2017 Math Help Forum. All rights reserved.