Correct and detailed answer will be Upvoted else downvoted. Thank you ### Problem Description**Task:** Cancel the cross product term from the quadratic form \[ 2x^2 + 10xy + 2y^2 \]### Detailed Explanation#### Step 1: Identify the Cross Product TermIn the given quadratic equation:\[ 2x^2 + 10xy + 2y^2 \]The term \( 10xy \) is the cross product term.#### Step 2: Transform the EquationTo cancel the cross product term, we can use a linear transformation. A common approach is to rotate the coordinate system by an angle \( \theta \) to eliminate the \( xy \) term. The transformed coordinates can be found using:\[ x = x' \cos \theta - y' \sin \theta \]\[ y = x' \sin \theta + y' \cos \theta \]#### Step 3: Calculate the AngleWe need to choose an angle \( \theta \) such that:\[ \tan(2\theta) = \frac{2 \cdot \text{coeff}(xy)}{\text{coeff}(x^2) - \text{coeff}(y^2)} = \frac{2 \cdot 10}{2 - 2} \]However, the denominator becomes zero in this case, indicating that we should rotate by \( \theta = \frac{\pi}{4} \) or 45 degrees.#### Step 4: Apply the TransformationWhen we apply the rotation \( \theta = 45^\circ \):\[ x = x' \cos 45^\circ - y' \sin 45^\circ \]\[ y = x' \sin 45^\circ + y' \cos 45^\circ \]Since \( \cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2} \), we get:\[ x = \frac{\sqrt{2}}{2} x' - \frac{\sqrt{2}}{2} y' \]\[ y = \frac{\sqrt{2}}{2} x' + \frac{\sqrt{2}}{2} y' \]By substituting these into the original quadratic form and simplifying, we would eliminate the cross product term.### ConclusionAfter applying a coordinate system rotation, the cross product term is effectively cancelled, simplifying the quadratic form to

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Answered: 2 Cen Cell the Crass preduct term from…

2 Cen Cell the Crass preduct term from the quardratee form 2

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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