# M 340L Exam 3 Solutions - Version 032 EXAM03 gilbert(56540...

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Unformatted text preview: Version 032 – EXAM03 – gilbert – (56540) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The graph of x T A x = 2 x 2 1 − 4 x 1 x 2 + 5 x 2 2 = 3 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. 1. straight lines 6 y 2 1 − y 2 2 = 0 2. ellipse y 2 1 + 6 y 2 2 = 3 correct 3. point (0 , 0) 4. hyperbola y 2 1 − y 2 2 = 3 5. parabola y 1 − 6 y 2 1 = 3 Explanation: The quadratic form can be written as 2 x 2 1 − 4 x 1 x 2 + 5 x 2 2 = [ x 1 x 2 ] bracketleftbigg 2 − 2 − 2 5 bracketrightbiggbracketleftbigg x 1 x 2 bracketrightbigg . The eigenvalues of A are the solutions of det bracketleftbigg 2 − λ − 2 − 2 5 − λ bracketrightbigg = λ 2 − 7 λ + 6 = ( λ − 1)( λ − 6) = 0 , i.e. , λ 1 = 1 and λ 2 = 6. Associated eigenvec- tors are u 1 = bracketleftbigg 2 1 bracketrightbigg , u 2 = bracketleftbigg − 1 2 bracketrightbigg ; these are orthogonal. The normalized eigen- vectors v 1 = 1 √ 5 bracketleftbigg 2 1 bracketrightbigg , v 2 = 1 √ 5 bracketleftbigg − 1 2 bracketrightbigg , are orthonormal, and P = [ v 1 v 2 ] = 1 √ 5 bracketleftbigg 2 1 − 1 2 bracketrightbigg is an orthogonal matrix such that A = P bracketleftbigg 1 6 bracketrightbigg 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 bracketleftbigg 1 6 bracketrightbigg y = y 2 1 + 6 y 2 2 = 3 . Consequently, the conic section is an ellipse . 002 10.0 points Find the solution of the differential equa- tion d u dt = A u ( t ) , u (0) = bracketleftbigg − 6 − 2 bracketrightbigg , when A is the matrix A = bracketleftbigg − 2 − 5 1 4 bracketrightbigg . 1. u ( t ) = bracketleftbigg 4 e- 3 t − 10 e t − 4 e- 3 t + 2 e t bracketrightbigg 2. u ( t ) = bracketleftbigg 4 e 3 t − 10 e- t − 4 e 3 t + 2 e- t bracketrightbigg correct 3. u ( t ) = bracketleftbigg − 4 e- 3 t + 10 e t 4 e 3 t − 2 e- t bracketrightbigg 4. u ( t ) = bracketleftbigg − 4 e 3 t + 10 e- t 4 e 3 t − 2 e- t bracketrightbigg Explanation: Version 032 – EXAM03 – gilbert – (56540) 2 Since det[ A − λI ] = vextendsingle vextendsingle vextendsingle vextendsingle − 2 − λ − 5 1 4 − λ vextendsingle vextendsingle vextendsingle vextendsingle = 5 − (2 + λ )(4 − λ ) = λ 2 − 2 λ − 3 = ( λ − 3)( λ + 1) , the eigenvalues of A are λ 1 = 3 , λ 2 = − 1 and corresponding eigenfunctions v 1 = bracketleftbigg 1 − 1 bracketrightbigg , v 2 = bracketleftbigg − 5 1 bracketrightbigg form a basis for R 2 because λ 1 negationslash = λ 2 . Thus u (0) = c 1 v 1 + c 2 v 2 ....
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