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@TuringTest pic to follow.

which problem are you doing?

all of them in exercise 6?

I assume you know to subtract lambda from the diagonals, then find the determinant of that matrix?

I'm doing 6E

I got [7*lambda-8+(lambda-5)*(lambda(lambda)-1)]

as shown in my pic

I don't think you simplified correctly...

Idk check my pic, that's what maple gave me anyways....

[(lambda-5)*(lambda-1)(lambda)+7*(0*0)-(-1)*(-1)]+[(lambda-1)*7+(lambda-5)*0-(0*(-1))*lambda]

I have always done det(A - lambda*I)=0

That's what the book shows us.

It comes from
\[ Ax =\lambda x\] or
\[ Ax -\lambda x=0\]
\[ (A -\lambda I) x=0\]

Yeah the book says that, then goes into normal determinants.e

Why is it only [7(λ−1)(−1)]?

the other parts have 0's so we can ignore them

and why do you only do it twice? I'm so confused. I thought the diagonal rule is 3 diags - 3 diags.

oic yeah.

deerrp :P

I don't like Maple's answer >( Wolfram doesn't either :P.

I use co-factors
(L-5) * det (lower right 2x2)
ignore the 0
-1 * det(lower left 2x2)

I hate co factors... From TT's example I get
[-7*lambda(-1)+1+(lambda(lambda-1))(lambda)] = 0

I just took what you did and entered it into Maple :P.

oh I think I know what Id id :P.

I frogot to add some extra brackets :). Does this shizzle look better >(

Same thing it looks like >(

TT never messes up :)

Normally maple gives me the good answer, but nope it's stupid :P.

yay :D
always trust your brain first is the moral, I'd say
Ok, dinner time, see ya!

NOOOOOOOOO!! 1 more quesitn plz.

Thanks, diff Q on the eigenvalues though :)

COME BACK SOON SO I CAN ASK MOAR Q'S!