▷ SECTION 5.2, QUESTION 4 2 OF 9 1 [100 A= 2 3 2 det (1²1-A) = det 020 Las 4 Loox L •)-3=0 EIGENVALUES X = 3 det (XI-A) = (1-3) 1-4 -1 (12) A-AI X=3 1 ro as O OO 9 1 2 1 4 ↑ 1-9-0 a = 9 4 F X+1=0 4 2 3 23 O [X-4 -1 ] = (2-3) [(x-4) (x-4)-(-1) (-2.5)] as X-4 (1-3) [ (x²-8) +1₁)-(25)] (1-3) (1²-81-9) (x-3)(x-2)(x+1) 1 2 4 X-40 1 →-2 1-3-2 -25
▷ SECTION 5.2, QUESTION 4 2 OF 9 1 [100 A= 2 3 2 det (1²1-A) = det 020 Las 4 Loox L •)-3=0 EIGENVALUES X = 3 det (XI-A) = (1-3) 1-4 -1 (12) A-AI X=3 1 ro as O OO 9 1 2 1 4 ↑ 1-9-0 a = 9 4 F X+1=0 4 2 3 23 O [X-4 -1 ] = (2-3) [(x-4) (x-4)-(-1) (-2.5)] as X-4 (1-3) [ (x²-8) +1₁)-(25)] (1-3) (1²-81-9) (x-3)(x-2)(x+1) 1 2 4 X-40 1 →-2 1-3-2 -25
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.3: Zeros Of Polynomials
Problem 54E
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I'm supposed to find the rank of the matrix for each eigenvalue.
I've found the eigenvalues and I know I'm supposed to find the eigenvectors, but I literally do not understand how. I've watched multiple videos, chatted with a tutor, and read the textbook, yet it still is not making sense. Can you show me how to find the eigenvector with detailed explanations 'Barney style' because I won't understand if you write the symobols you see in proofs and textbooks.
I've attached what I have completed so far.
![→
SECTION 5.2, QUESTION
2 OF 9
4
1
1100
4 O 1
> A = 2
2 det (11-A) = det oro - 23
оло
▷
4
oox
23
3
Las o
▷
▷ • λ-3=0
EIGENVALUES X = 3
▷
DA-AI
X= }
1
▷
► det (XI-A) = (x-3) [ 1-4 - ] = (1-3) [(x-4) (x-4)-(-1)(-25)]
X-4] (x-3) [ (1²-8) +16) - (25)]
(1-3) (1²-81-9)
(x-3)(x-9) (x+1)
O
2
9
Las o
1
2
1
a
↑
->
X-9=0
=9
as x4
X+1=0
(= -1
9
O
0
2
4
1-4 -1
→-2 1-3 -2
L-25 0
X-4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48ae96ba-8296-4f48-84d9-a65ba8f4727e%2Fad4197b1-3162-4e09-bc00-01bd659d0fa9%2Fltd3xtf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:→
SECTION 5.2, QUESTION
2 OF 9
4
1
1100
4 O 1
> A = 2
2 det (11-A) = det oro - 23
оло
▷
4
oox
23
3
Las o
▷
▷ • λ-3=0
EIGENVALUES X = 3
▷
DA-AI
X= }
1
▷
► det (XI-A) = (x-3) [ 1-4 - ] = (1-3) [(x-4) (x-4)-(-1)(-25)]
X-4] (x-3) [ (1²-8) +16) - (25)]
(1-3) (1²-81-9)
(x-3)(x-9) (x+1)
O
2
9
Las o
1
2
1
a
↑
->
X-9=0
=9
as x4
X+1=0
(= -1
9
O
0
2
4
1-4 -1
→-2 1-3 -2
L-25 0
X-4
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