Eigenvalues and eigenvectors, 3x3 matrix using Remainder and Factor Theorem

AI Thread Summary
The discussion focuses on finding the eigenvalues of a 3x3 matrix using the Remainder and Factor Theorem. The user attempts to calculate the determinant and factor the resulting polynomial but encounters confusion regarding their calculations. They specifically question whether they should have factored out (0-λ) and whether their simplifications were correct. Ultimately, they seek clarification on solving the quadratic equation λ² - 2λ + 1 = 0 to find the roots. The conversation emphasizes the importance of correctly applying algebraic techniques to determine eigenvalues.
Jowin86
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Homework Statement



i = the 3x3 matrix below

2-λ 0 1
-1 4-λ -1
-1 2 0-λ

Using remainder and factor theorem find the 3 values of λ.

Homework Equations



|i| = a1|b2c3-c2b3|-a2|a2c3-c2a3|+a3|a2b3-b2a3|

|a|=ad-bc

The Attempt at a Solution

(2-λ) |(4-λ x 0-λ)-(-1x2)|+1|(-1x2)-(4-λ x -1) **because b1 is 0 I've left it out**

(2-λ)[(4-λ)(0-λ)+2] +1 [-2+(4-λ)]

(2-λ)(4-λ)(0-λ)+1(4-λ)

Factorise out (4-λ):

(4-λ)[(2-λ)(0-λ)+1]

Multiply [] brackets out (FOIL):

(4-λ)(λ2-2λ+1)

...and now I'm stuck which probably means I've gona wrong somewhere ;-(

*should I have taken out 0-λ?

*on the 2nd line of my attempted answer I figured the +2 in the first square brackets canceled out the -2 in the second square brackets, was I wrong?Thanks for any hints and help :)
 
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2-2λ+1) can be factored.

If you can't do that, then find the roots: λ2-2λ+1 = 0
 
(4-λ) λ(-2λ)

λ=4 λ=2 λ=0

Is that wrong?

I'm not sure how to find the roots of λ2-2λ+1=0 ?
 
Jowin86 said:
(4-λ) λ(-2λ)

λ=4 λ=2 λ=0

Is that wrong?
Yes, it's wrong
I'm not sure how to find the roots of λ2-2λ+1=0 ?

It's a quadratic equation. Solve it.
 
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