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Unformatted text preview: Math 221, Solutions for assignment due Nov. 11 Section 7.2 Problem 10 det 3 4 1  2 7 3 = ( 3 )( 1 )(3 ) 2( 4( 1 )) = ( +1) 2 (  1) So the matrix has eigenvalues = 1, with algebraic multiplicity 1 and = 1, with algebraic multiplicity 2. Problem 12 det 2  2 1 1 3  4 2 3 = det bracketleftbigg 2  2 1 1 bracketrightbigg det bracketleftbigg 3  4 2 3 bracketrightbigg = ( 2 )( 2 1) = ( + 1)(  1) 2 . So The eigenvalues are 0 , 1 (with algebraic multiplicity 1) and 1 (with algebraic multi plicity 2). Problem 19 This is true. The characteristip polynomial is 2 ( trace )  det with det < 0, so the discriminant of this polynomial is D = trace 2 4 det . Since the term under the square root sign is strictly positive, the characteristic polynomial has two distinct roots: trace + D 2 and trace D 2 . Problem 22 det( A I ) = det( A I ) T = det( A T I ). Thus the characteristic polynomials of A and A T are the same, which implies that the eigenvalues are the same. Problem 22 If B = S 1 AS , then det( A I ) = det S 1 ( A I ) S = det( S 1 AS S 1 IS ) = det( S 1 AS I ) = det( B I ) ....
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This homework help was uploaded on 02/23/2008 for the course MATH 2210 taught by Professor Pantano during the Fall '05 term at Cornell University (Engineering School).
 Fall '05
 PANTANO
 Linear Algebra, Algebra, Equations

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