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?

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- May 17th 2009, 09:19 PMscorpion007Possibly 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 PMJen
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 PMGammaCofactor 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 AMscorpion007
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 AMmr fantastic
The equation can be written as

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

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

Factorise this by taking out the common factor and simplifying.

Note: The final answer given in your image is wrong. - May 18th 2009, 03:32 AMscorpion007
Ah thank you!

So it should be:

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

These lecture notes have many mistakes...

edit:

which is the same as:

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

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

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

and so

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