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Unformatted text preview: Math 2J Lecture B Fall 2009 (Instructor: Professor R. C. Reilly) Answer Key for Practice Midterm #2 (1) (a) Iâ€™ll use the direct formula p ( Î» ) = Î» 3 +( a 11 + a 22 + a 33 ) Î» 2 ( A 11 + A 22 + A 33 ) Î» +det( A ) that was discussed in class. By inspecting the entries of the matrix A = 4 2 2 1 3 1 1 1 1 , one sees that a 11 + a 22 + a 33 = 4 + 3 + 1 = 8; A 11 + A 22 + A 33 = ( 1) 1+1 3 1 1 1 + ( 1) 2+2 4 2 1 1 + ( 1) 3+3 4 2 1 3 = 4 + 6 + 10 = 20 . Finally, by using the cofactor expansion along the first row, one gets det( A ) = 4 A 11 + 2 A 12 2 A 13 = 16 4 + 4 = 16 . Thus, the characteristic polynomial is p ( Î» ) = Î» 3 + 8 Î» 2 20 Î» + 16 . Remark The problem does not require you to factor this polynomial. If you did the factorization, however, and came up with â€˜eigenvaluesâ€™ other than the ones the problem tells you are the only ones, you should have figured out that you made a mistake! (b) The Case Î» 1 = 2 One computes A 2 I = 2 2 2 1 1 1 1 1 1 â†’ 2 2 2 0 0 0 0 â†’ 1 1 1 0 0 0 0 The first step in the GaussJordan process subtracts R 1 / 2 from R 2 and R 3 ; the second step divides R 1 by 2. The final matrix is in reduced rowechelon form, and it shows that the general solution x of the homogeneous system has two free variables, namely x 2 and x 3 . The formula is x = x 2  1 1 + x 3 1 1 Thus, the geometric multiplicity of the eigenvalue Î» 1 = 2 for A is 2. The Case Î» 2 = 4 Now one computes A 4 I = 2 2 1 1 1 1 1 3 â†’ 1 1 1 2 2 1 1 3 â†’ 1 1 1 2 2 2 2 â†’ 1 1 1 2 2 â†’ 1 1 1 1 1 â†’ 1 0 2 0 1 1 0 0 From this last matrix, which is in reduced rowechelon form, one reads off that the eigenspace for the eigenvalue Î» 2 = 4 consists of all vectors x of the form x = x 3 2 1 1 In particular, the geometric multiplicity of Î» 2 equals 1.equals 1....
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 Spring '10
 DONALDSON,NEIL
 Taylor Series, Taylor Polynomials, Eigenvalue, eigenvector and eigenspace, Natural logarithm, general solution

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