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solmidterm2

# solmidterm2 - Harvard University Solutions Midterm 2 for...

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Harvard University Solutions Midterm 2 for Math 121, Fall 2007 Wednesday, November 21, 2007 1. (20 points) a) False b) False c) False d) False e) True 2. (50 points) a) (30 points) A = 2 2 3 1 2 1 2 - 2 1 First, we compute the characteristic polynomial p A ( λ ). p A ( λ ) = det ( A - λI ) = 2 - λ 2 3 1 2 - λ 1 2 - 2 1 - λ = - λ 3 +5 λ 2 - 2 λ - 8 = - ( λ +1)( λ - 2)( λ - 4) Thus, it is det ( A ) = p A (0) = - 8 Therefore, A is invertible and rank ( A ) = 3. The eigenvalues are the zeros of the charac- teristic polynomial p A ( λ ). That is they are the solutions of the equation p A ( λ ) = 0 Using the hint that ( - 1) is an eigenvalue of A , one can easily factorize p A ( λ ) as above. We obtain the eigenvalues of A being λ 1 = - 1, λ 2 = 2, λ 3 = 4. The eigenvectors then are computed as nonzero solutions of ( A - λ i I ) v = 0 for i = 1 , 2 , 3 . Here, the eigenvector corresponding to λ 1 has already been given in the hint. We find the eigenvectors v 1 = (1 , 0 , - 1) corresponding to λ 1 = - 1 v 2 = (2 , 3 , - 2) corresponding to λ 2 = 2 v 3 = (8 , 5 , 2) corresponding to λ 3 = 4.

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solmidterm2 - Harvard University Solutions Midterm 2 for...

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