# Find the eigen value

## The Attempt at a Solution

Do you see where it says 3.3(1-lambda)?

Then the next row has -9 in it? It should be -9 + 9lambda, right? I'm using the Sars method (I think that's what it's called), where you copy the first two columns then take the product of three different diagonals, then subtract them from 3 other diagonals, sorry for the vague language.

## Answers and Replies

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Hi robert!

Only $1-\lambda$ was taken common of the two terms, and the rest was kept as it is. I dont see why you would have a $+9\lambda$ with it.

Let $(1-\lambda) = t$ and keep the rest as it is. And take the t out common, what would you get??

i'm falling asleep, so I will have to deal with this 10 hours later.

i'm falling asleep, so I will have to deal with this 10 hours later.
Good idea!! :tongue:

HallsofIvy
Homework Helper

## The Attempt at a Solution

Do you see where it says 3.3(1-lambda)?

Then the next row has -9 in it? It should be -9 + 9lambda, right?
No. They have factored $1- \lambda$ from each term first, leaving $(2- \lambda)(2-\lambda)- 3(3)$.

I'm using the Sars method (I think that's what it's called), where you copy the first two columns then take the product of three different diagonals, then subtract them from 3 other diagonals, sorry for the vague language.
Not a very good method as it is applicable only to 3 by 3 determinants and not to larger determinants. Better to learn "expansion by rows" or "expansion by columns".

Here, it is particularly easy to expand on the last row. Then the determinant is $$0\left|\begin{array}{cc}3 & 0 \\ 2-\lambda & 0 \end{array}\right|- 0\left|\begin{array}{cc}2-\lambda & 0 \\ 3 & 0 \end{array}\right|+ (1- \lambda)\left|\begin{array}{cc}2-\lambda & 2 \\ 3 & 2- \lambda\end{array}\right|$$
which, since the first two terms are multiplied by 0, immediately gives $((2-\lambda)^2- 9)(1- \lambda)$.

ok, I understand now.