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Unformatted text preview: 4b 6 0 3a 5b 8 0 We can use any two of these to solve for a and b to get a=‐1, b=‐1. 6. a. Again using the formula that trace equals sum of eigenvalues, we get that the other eigenvalue of A must be 4. b. We know that the determinant is the product of eigenvalues and must therefore be 4. On the other hand, det(A)=6‐b2. Therefore, b2=2 and b= 2. 7. Let’s solve B part first. We are told that A is an nxn matrix whose characteristic polynomial is s n n 1s n 1 n 2 s n 2 1s 0 We start by writing characteristic polynomial of 2A: s s det( sI 2 A) det 2 I A 2n det I A . 2 2 Let us define a new variable z=s...
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This note was uploaded on 12/27/2011 for the course EEL 3105 taught by Professor Boykins during the Fall '10 term at University of Florida.

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