# Possibly a determinant question

• May 17th 2009, 09:19 PM
scorpion007
Possibly a determinant question
In the following image,
http://img33.imageshack.us/img33/9210/math7.png

How did they get from step (1) to (2)?

What does "Expanding across the first row" mean here?
• May 17th 2009, 09:49 PM
Jen
Take the determinant.

Expanding across the first row mean they are taking the determinant starting in the top left corner with the first element of the first row.
• May 17th 2009, 10:24 PM
Gamma
Cofactor expansion.
This is called cofactor expansion. You can calculate the determinant by going down any column (the determinant of the transpose is the same, so you can also go down a row if you want.) Determinant Expansion by Minors -- from Wolfram MathWorld
• May 18th 2009, 12:12 AM
scorpion007
Hopefully this isn't breaking the rule of two unrelated questions in a single thread, but it relates to the same image:

How did they factorise the cubic polynomial into those two factors? Does it involve polynomial division? I can't seem to see it.
• May 18th 2009, 03:13 AM
mr fantastic
Quote:

Originally Posted by scorpion007
Hopefully this isn't breaking the rule of two unrelated questions in a single thread, but it relates to the same image:

How did they factorise the cubic polynomial into those two factors? Does it involve polynomial division? I can't seem to see it.

The equation can be written as

$-\lambda (\lambda + 1) (\lambda - 1) + (\lambda + 1) + (\lambda + 1) = 0$

$-\lambda (\lambda + 1) (\lambda - 1) + 2(\lambda + 1) = 0$

Factorise this by taking out the common factor and simplifying.

• May 18th 2009, 03:32 AM
scorpion007
Ah thank you!

So it should be:

$(\lambda +1)(-\lambda^2+\lambda+2)=0$ ?

These lecture notes have many mistakes...

edit:
which is the same as:

$-(\lambda +1)(\lambda^2-\lambda-2)=0$

$\implies -(\lambda +1)(\lambda-2)(\lambda+1)=0$

$\implies -(\lambda +1)^2(\lambda-2)=0$

and so

$\lambda = -1, 2$ (has a repeated root).