# EXAM03-solutions - Version 047 EXAM03 gilbert(55730 This...

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Version 047 – EXAM03 – gilbert – (55730) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Make an orthogonal change oF variables that reduces x T A x = 2 x 2 1 + 12 x 1 x 2 11 x 2 2 = 5 to a quadratic equation in y 1 , y 2 with no cross- product term given that λ 1 λ 2 . 1. y 2 1 14 y 2 2 = 5 2. y 2 1 + 14 y 2 2 = 5 3. y 2 1 14 y 2 2 = 5 correct 4. y 2 1 + 14 y 2 2 = 5 Explanation: In matrix terms, x T A x = 2 x 2 1 + 12 x 1 x 2 11 x 2 2 = [ x 1 x 2 ] b 2 6 6 11 Bb x 1 x 2 B . The eigenvalues λ 1 , λ 2 oF A are the solutions oF det pb 2 λ 6 6 11 λ BP = λ 2 + 13 λ 14 = ( λ 1)( λ + 14) = 0 , i.e. , λ 1 = 1 and λ 2 = 14. Associated eigen- vectors are u 1 = b 2 1 B , u 2 = b 1 2 B , and these are orthogonal since λ 1 n = λ 2 . The normalized eigenvectors v 1 = 1 5 b 2 1 B , v 2 = 1 5 b 1 2 B , are thus orthonormal, and P = [ v 1 v 2 ] = 1 5 b 2 1 1 2 B is an orthogonal matrix such that A = P b 1 0 0 14 B P - 1 = PDP T is an orthogonal diagonalization oF A . Now set x = b x 1 x 2 B , y = b y 1 y 2 B , x = P y . Then x T A x = ( P y ) T ( PDP T ) P y = y T ( P T P ) D ( P T P ) y = y T D y = y T b 1 0 0 14 B y = y 2 1 14 y 2 2 = 5 . Consequently, when x = P y , 2 x 2 1 + 12 x 1 x 2 11 x 2 2 = y 2 1 14 y 2 2 = 5 . 002 10.0 points IF { x 1 , x 2 , x 3 } is a linearly independent set and W = Span { x 1 , x 2 , x 3 } , then any orthogonal set { v 1 , v 2 , v 3 } in W is a basis For W . True or ±alse? 1. ±ALSE correct 2. TRUE Explanation: The three orthogonal vectors must be non- zero to be a basis For a three-dimensional subspace. Consequently, the statement is incomplete and thus ±ALSE . 003 10.0 points
Version 047 – EXAM03 – gilbert – (55730) 2 If A is symmetric, then the change of vari- able x = P y transforms Q ( x ) = x T A x into a quadratic form with no cross-product term for any orthogonal matrix P . True or False? 1. FALSE correct 2. TRUE Explanation: When P is orthogonal and x = P y , then Q ( x ) = x T A x = ( P y ) T A ( P y ) = y T P T AP y = y T ( P T AP ) y . But this will contain cross-product terms un- less P T AP is a diagonal matrix, i.e. , unless A is orthogonally diagonalized by P . Consequently, the statement is FALSE . 004 10.0 points When u 1 = b 1 2 B , u 2 = b 2 1 B , are eigenvectors of a symmetric 2 × 2 matrix A corresponding to eigenvalues λ 1 = 2 , λ 2 = 3 , ±nd matrices D and P in an orthogonal diag- onalization of A . 1. D = b 2 0 0 3 B , P = 1 5 b 1 2 2 1 B correct 2. D = b 3 0 0 2 B , P = 1 5 b 1 2 2 1 B 3. D = b 2 0 0 3 B , P = b 1 2 2 1 B 4. D = b 2 0 0 3 B , P = 1 5 b 2 1 1 2 B 5. D = b 3 0 0 2 B , P = b 1 2 2 1 B 6. D = b 2 0 0 3 B , P = b 2 1 1 2 B Explanation: When D = b λ 1 0 0 λ B , Q = [ u 1 u 2 ] , then Q has orthogonal columns and A = QDQ - 1 is a diagonalization of A , but it is
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