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Unformatted text preview: Solutions  Midterm II J. Scholtz December 4, 2006 Question 1 1. Let us nd the characteristic polynomial rst since that gives us majority of the information we will need: p ( t ) = det( AtI ) = 1t 3 2 4t 2 3 2 1t =t 3 + 6 t 2 + 4 t24 =( t 36 t 24 t + 24) This means that det( A ) = p (0) =24. Hence A is invertible and its rank then must be 3. We know that the eigenvalues are the roots of the characteristic polynomial. There are couple of ways to factorize it, but since we were given that the roots are integers then we could just nd one by trial and error and divide the two polynomials. On the other hand from algebra we know that the second leading term of the polynomial gives us the negativ eof sume of roots and the constant term is their negative poduct in this case hence abc =24 and a + b + c = 6, a, b, c Z . The only solution to this is a = 2, b =2 and c = 6  the eigenvalues....
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This note was uploaded on 07/19/2008 for the course MATH 110 taught by Professor Gurevitch during the Summer '08 term at University of California, Berkeley.
 Summer '08
 GUREVITCH
 Linear Algebra, Algebra

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