determinant

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• Aug 19th 2007, 06:52 AM
needhelpsobad
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.