Question

Find all the eigenvalues for the given matrices.
if it is a 3x3. and a polynomial. how do i know what to break it down into?

Answer 1

[ 4 0 1
-1 -6 -2
5 0 0 ]

\[\lambda ^3+2 lambda ^2 - 29 lambda -30]/

Answer 2

\[\textbf Av=\lambda v\]
\[\left(\textbf A-\lambda \textbf I\right)v =0\]
\[\left|\textbf A-\lambda \textbf I\right|v =0\]
\[\left|\left(\begin{array}\ 4&0&1\\-1&-6&-2\\5&0&0\end{array}\right)-\left(\begin{array}\ \lambda&0&0\\0&\lambda & 0\\0 &0 &\lambda \end{array}\right)\right|=0\]
\[\left|\left(\begin{array}\ 4-\lambda&0&1\\-1&-6-\lambda&-2\\5&0&0-\lambda\end{array}\right)\right|=0\]

\[(4-\lambda)[-2\times0-(-6-\lambda)(0-\lambda)]-0[-1\times(0-\lambda)]+1[-1\times0-(-2\times5)]=0\]

Answer 3

*\[(4-\lambda)[-2\times0-(-6-\lambda)(0-\lambda)]\]\[\qquad-0[-1\times(0-\lambda)-(5\times-2)]\]\[\qquad\qquad+1[-1\times0-(-6-\lambda)]=0\]

Answer 4

right, i know how to do all the calculations but when you get the the last step: when its all combined how do you know what lamda itself is equal to? ( like, you look at the constant and see its factors or something and do something..idk)_)

Answer 5

well at the last spet you'll have a bunch of factors =0

so to get zero at least one of the factors is zero

Answer 6

something like this only i have made mistakes \[(\lambda-4)[(6+\lambda)\lambda]+[6+\lambda]=0\]\[(6\lambda+\lambda^2-24-4\lambda)\lambda+6+\lambda=0\]\[\lambda^3+2\lambda^2-24\lambda+\lambda+6=0\]\[\lambda^3+3\lambda^2-24\lambda+6=0\]\[(\lambda-a)(\lambda-b)(\lambda-c)=0\]

lambda would be a,b,c

Answer 7

but how do you know to break it into a b c?

Answer 8

how to factorize?
you could try polynomial long division

Answer 9

ugh

Answer 10

\[\lambda^3 +2\lambda^2-29\lambda-30\]

Answer 11

how would you go about doing this?
possibilities: +/- : 1,2,3,5,6,10,30,15

Answer 12

if you plug in -1, you get =0.then what?

Answer 13

your expression does factorizes nicely,
try dividing by \((\lambda+1)\)

Answer 14

i dont know how... i knwo that sounds dumb but how do u do that?