# But this will contain cross product terms un less p t

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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 . 007 10.0 points If A is a 2 × 2 symmetric matrix, then the set of x such that x T A x = c , for some constant c , corresponds to either a circle, an ellipse, or a hyperbola. True or False? 1. TRUE
jones (dzj62) – HW14 – gilbert – (53525) 3 2. FALSE correct Explanation: There are also “degenerate” cases where the solution set of x T A x = c can be a single point, two intersecting lines, or no points at all. Consequently, the statement is FALSE . 008 10.0 points A quadratic form can always be written as Q ( x ) = x T A x with A a symmetric matrix. True or False? 1. TRUE correct 2. FALSE Explanation: By definition a quadratic form has the form Q ( x ) = a 11 x 2 1 + . . . + a nn x 2 n + 2( a 12 x 1 x 2 + . . . + a n 1 ,n x n 1 x n ) . Thus Q ( x ) = x T A x with x = x 1 x 2 . . . x n , A = a 11 a 12 . . . a 1 n a 12 a 22 . . . a 2 n . . . . . . . . . . . . a 1 n a 2 n . . . a nn , and A T = A . Consequently, the statement is TRUE . 009 10.0 points The graph of x T A x = x 2 1 12 x 1 x 2 + 15 x 2 2 = 1 is a conic section. Find an orthogonal matrix P so that the transformation x = P y diagonalizes A , and then use this to identify the conic section given that λ 1 λ 2 . 1. ellipse 17 y 2 1 3 y 2 2 = 1 2. ellipse 17 y 2 1 + 3 y 2 2 = 1 3. hyperbola 17 y 2 1 3 y 2 2 = 1 correct 4. hyperbola 17 y 2 1 + 3 y 2 2 = 1 5. empty set 6. point (0 , 0) 7. straight lines 17 y 2 1 3 y 2 2 = 0 Explanation: The quadratic form can be written as x 2 1 12 x 1 x 2 + 15 x 2 2 = [ x 1 x 2 ] 1 6 6 15 x 1 x 2 . The eigenvalues of A are the solutions of det 1 λ 6 6 15 λ = λ 2 14 λ 51 = ( λ 17)( λ + 3) = 0 , i.e. , λ 1 = 17 and λ 2 = 3. Associated eigen- vectors are u 1 = 1 3 , u 2 = 3 1 ; these are orthogonal. The normalized eigen- vectors v 1 = 1 10 1 3 , v 2 = 1 10 3 1 , are orthonormal, and P = [ v 1 v 2 ] = 1 10 1 3 3 1
jones (dzj62) – HW14 – gilbert – (53525) 4 is an orthogonal matrix such that A = P 17 0 0 3 P 1 = PDP T is an orthogonal diagonalization of A . Now set 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 17 0 0 3 y = 17 y 2 1 3 y 2 2 = 1 . Consequently, the conic section is a hyperbola . 010 10.0 points The graph of x T A x = 3 x 2 1 + 12 x 1 x 2 + 19 x 2 2 = 5 is a conic section. Find an orthogonal matrix P so that the transformation x = P y diagonalizes A , and then use this to identify the conic section given that λ 1 λ 2 . 1. straight lines 21 y 2 1 + y 2 2 = 0 2. hyperbola 21 y 2 1 y 2 2 = 5 3. ellipse 21 y 2 1 + y 2 2 = 5 correct 4. ellipse 21 y 2 1 y 2 2 = 5 5. empty set 6. hyperbola 21 y 2 1 + y 2 2 = 5 7. point (0 , 0) Explanation: The quadratic form can be written as 3 x 2 1 + 12 x 1 x 2 + 19 x 2 2 = [ x 1 x 2 ] 3 6 6 19 x 1 x 2 . The eigenvalues of A are the solutions of det 3 λ 6 6 19 λ = λ 2 22 λ + 21 = ( λ 21)( λ 1) = 0 , i.e. , λ 1 = 21 and λ 2 = 1. Associated eigen- vectors are u 1 = 1 3 , u 2 = 3 1 ; these are orthogonal. The normalized eigen- vectors v 1 = 1 10 1 3 , v 2 = 1 10 3 1 , are orthonormal, and P = [ v 1 v 2 ] = 1 10 1 3 3 1
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