determinant

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Aug 19th 2007, 06:52 AM
needhelpsobad

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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.

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