Let Q(x) = 143x² – 34x2 – 34x² – 36x1x2 – 36x1x3 — 72x2x3. Find a unit vector x in R3 whose first entry is positive at which Q(x) is minimized, subject to x¹x = 1. [Hint: The eigenvalues of the matrix of the quadratic form Q are 2, -73, and 146.] Enter the components of x (in order) into the answer box below, separated with commas.
Let Q(x) = 143x² – 34x2 – 34x² – 36x1x2 – 36x1x3 — 72x2x3. Find a unit vector x in R3 whose first entry is positive at which Q(x) is minimized, subject to x¹x = 1. [Hint: The eigenvalues of the matrix of the quadratic form Q are 2, -73, and 146.] Enter the components of x (in order) into the answer box below, separated with commas.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.6: The Algebra Of Matrices
Problem 13E
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![Let
Q(x) = 143x² – 34x2 – 34x² – 36x1x2 – 36x1x3 – 72x2x3.
-
-
T
Find a unit vector x in R³ whose first entry is positive at which Q(x) is minimized, subject to x¹x = 1. [Hint: The
eigenvalues of the matrix of the quadratic form Q are 2, -73, and 146.]
Enter the components of x (in order) into the answer box below, separated with commas.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff316efbb-9e65-469e-bdde-f3c49a257ac3%2F7da785bc-73a4-4cc8-891b-fd9e1e285621%2Fuabxppm_processed.png&w=3840&q=75)
Transcribed Image Text:Let
Q(x) = 143x² – 34x2 – 34x² – 36x1x2 – 36x1x3 – 72x2x3.
-
-
T
Find a unit vector x in R³ whose first entry is positive at which Q(x) is minimized, subject to x¹x = 1. [Hint: The
eigenvalues of the matrix of the quadratic form Q are 2, -73, and 146.]
Enter the components of x (in order) into the answer box below, separated with commas.
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