Question

I need help finding the Eigenvalue for a specific matrix.

Answer 1

Watch matrix

Answer 2

Sorry I couldnt connect to the server...

Answer 3

@ganeshie8 Please explain how we go about finding the eigenvalue of this type of matrix...because this is not a typical 3x3 matrix there something different about it.

Answer 4

@freckles

Answer 5

something different about it?
can't you just apply the characteristic equation?
that is solve:
\[det(A-\lambda \cdot I)=0\]

Answer 6

Ya I did that but it was not right

Answer 7

\[\text{ Example } \\ \text{ Let } A=\left[\begin{matrix}2 & -3 & 4\\5 & -3 & -1 \\ -4 & 4 & 1\end{matrix}\right] \\ \text{ then } A-\lambda \cdot I=\left[\begin{matrix}2-\lambda & -3 & 4\\5 & -3-\lambda & -1 \\ -4 & 4 & 1-\lambda\end{matrix}\right] \\ \\ =(2- \lambda)[(-3-\lambda)(1-\lambda)-(-4)]-(-3)[5(1-\lambda)-(-1)(-4)] \\ +4[5(4)-(-3-\lambda)(-4)]\]
then you set =0 and solve for lambda

Answer 8

I did \[\det(\lambda I-A)=0\]

Answer 9

did you get the equation:
\[(-6-\lambda)[(-3-\lambda)(9-\lambda)-(-12)(3)] \\ -(-3)[-12(9-\lambda)-(-12)(9)] \\ +(-6)[-12(3)-9(-3-\lambda)]=0\]

Answer 10

what did you get for x or lambda

Answer 11

ok let me try this one more time

Answer 12

err if I didn't make a type-0 above it should
simplify to:
\[9x-x^3=0 \text{ or } x^3-9x=0\]

Answer 13

oh woops I also made a mistake on my paper...But yea I got that too now...Before I was multiplying (x+6) by the whole thing, which caused it to be wrong.

Answer 14

Thank you

Answer 15

np