Determining the three values of λ

[Cameron Blink]

Hi , guys and hope you've had a great week till now.

I have mathematics 1 exam coming up reaaaally soon (28th of april) and i am struggling on the MOCK exam. InitiallyI've skipped the first question, even though it would give the most points, but since i started to getting stuckon the other questions, I went back on the first one. And guess what...i am stuck here as well xD.
My personal tutor isn't responding to the mails I've sent him AND on top of that our lecturer does not bother to do an online-consultations. How am i suppose to practise for the exam 24/7 if nobody bothers to help me. I am uploadinga picture of the first question.

http://store.picbg.net/pubpic/E0/6C/4c56b407e286e06c.jpg - screenshot

Note* :
I don't ask to solve it for me , can you try to teach me how to do it on my own.
Thank you.

 

[HallsofIvy]

In other words, you are asked to find the eigenvalues or A. Since that involves solving an equation based on a determinant, I assume you know how to evaluate determinants.

Seeing those two "0"s in the first row I would consider expand0ting the determinant on that first row,

\(\displaystyle \left|\begin{array}{ccc}-\lambda & 0 & x \\ a & 5- \lambda & y \\ c & d & 5- \lambda \end{array}\right|= -\lambda\left|\begin{array}{cc} 5-\lambda & y \\ d & 5-\lambda \end{array}\right|+ x\left|\begin{array}{cc}a & 5-\lambda \\ c & d \end{array}\right|
= -\lambda[(5-\lambda)^2- dy]+ x[ad- c(5-\lambda)]= 0
\)
That is a cubic equation to solve for \(\displaystyle \lambda\). Do not forget the given:

ad = 5c................and

cx + dy = 9