# Thus q x x t a x with x x 1 x 2 x n a a 11 a 12 a 1 n

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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 . 011 0.0 points The graph of x T A x = x 2 1 12 x 1 x 2 + 17 x 2 2 = 2 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. point (0 , 0) 2. empty set
farias (df8272) – ReviewHW14 – gilbert – (56780) 5 3. ellipse 19 y 2 1 y 2 2 = 2 4. hyperbola 19 y 2 1 y 2 2 = 2 correct 5. ellipse 19 y 2 1 + y 2 2 = 2 6. hyperbola 19 y 2 1 + y 2 2 = 2 7. straight lines 19 y 2 1 y 2 2 = 0 Explanation: The quadratic form can be written as x 2 1 12 x 1 x 2 + 17 x 2 2 = [ x 1 x 2 ] 1 6 6 17 x 1 x 2 . The eigenvalues of A are the solutions of det 1 λ 6 6 17 λ = λ 2 18 λ 19 = ( λ 19)( λ + 1) = 0 , i.e. , λ 1 = 19 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 is an orthogonal matrix such that A = P 19 0 0 1 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 19 0 0 1 y = 19 y 2 1 y 2 2 = 2 . Consequently, the conic section is a hyperbola . 012 0.0 points The graph of x T A x = 2 x 2 1 + 12 x 1 x 2 + 14 x 2 2 = 4 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 16 y 2 1 4 y 2 2 = 4 2. point (0 , 0) 3. hyperbola 16 y 2 1 4 y 2 2 = 4 correct 4. ellipse 16 y 2 1 + 4 y 2 2 = 4 5. empty set 6. straight lines 16 y 2 1 4 y 2 2 = 0 7. hyperbola 16 y 2 1 + 4 y 2 2 = 4 Explanation: The quadratic form can be written as 2 x 2 1 + 12 x 1 x 2 + 14 x 2 2 = [ x 1 x 2 ] 2 6 6 14 x 1 x 2 . The eigenvalues of A are the solutions of det 2 λ 6 6 14 λ = λ 2 12 λ 64 = ( λ 16)( λ + 4) = 0 ,
farias (df8272) – ReviewHW14 – gilbert – (56780) 6 i.e. , λ 1 = 16 and λ 2 = 4. 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 is an orthogonal matrix such that A = P 16 0 0 4 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 16 0 0 4 y = 16 y 2 1 4 y 2 2 = 4 . Consequently, the conic section is a hyperbola . 013 0.0 points Use an orthogonal diagonalization A = P λ 1 0 0 0 λ 2 0 0 0 λ 3 P T with λ 1 > λ 2 > λ 3 to reduce x T A x = 3 x 2 1 + x 2 2 + x 2 3 + 4 x 1 x 2 + 4 x 1 x 3 + 8 x 2 x 3 = 2 to a quadratic equation in y 1 , y 2 , y 3 with no cross-product terms.