Suppose Q is the quadratic form below The minimum value of Q subject to = 1 is Q = 4. An eigenvector of A associated with eigenvalue λ = 4 is What is c equal to? -2 An eigenvector associated with eigenvalue λ = 4 is = (0, 1, -2). The minimum value of Q, subject to #= 1 is obtained at Zo, where: If is parallel to and k > 0, what must k₁ be equal to? (answer must contain at least 3 decimal places) -2 /5 2 1 Q = ¹ AZ, A = 2 8 2 1 2 5, -- (1) = to =

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter5: Orthogonality
Section5.3: The Gram-schmidt Process And The Qr Factorization
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Suppose Q is the quadratic form below
The minimum value of Q subject to ¹ = 1 is Q = 4. An eigenvector of A associated with eigenvalue λ = 4 is
What is c equal to? -2
An eigenvector associated with eigenvalue λ = 4 is ₹ = (0, 1, −2)². The minimum value of Q, subject to ª € = 1 is obtained at to, where:
If 7 is parallel to to and ko > 0, what must k₁ be equal to? (answer must contain at least 3 decimal places) -2
Q = ¹ Añ, A = | 2
V =
To
5 2 1
-(:)
8 2
1 2 5
||
(1)
ko
Transcribed Image Text:Suppose Q is the quadratic form below The minimum value of Q subject to ¹ = 1 is Q = 4. An eigenvector of A associated with eigenvalue λ = 4 is What is c equal to? -2 An eigenvector associated with eigenvalue λ = 4 is ₹ = (0, 1, −2)². The minimum value of Q, subject to ª € = 1 is obtained at to, where: If 7 is parallel to to and ko > 0, what must k₁ be equal to? (answer must contain at least 3 decimal places) -2 Q = ¹ Añ, A = | 2 V = To 5 2 1 -(:) 8 2 1 2 5 || (1) ko
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I should have noted that I also tried -2.000, it is also incorrect. 

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The image displays that K1 = -2 is wrong. I got the same answer in my own work and still confused on what I'm doing wrong. 

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