EEL3105_MT2_Solution

EEL3105_MT2_Solution - /Fall2011 14 6 28 6 1.a...

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EEL 3105 Midterm 2 Solution /Fall 2011 1. a. 5 13 6 14 AB , (other version 5 26 6 28 AB ) b. 9 3 3 6 13 4 AB c. AB is not defined as the number of columns of A does not equal the number of rows of B. 2. For A to be singular, det(A). We can compute det(A) by expanding using the 3 rd row to give det( ) 1.(4 ( 4)) (2 ( 2)) 8 4 0 A   [The determinant of A for the other version is 4 2 .] Therefore, α =2. 3. a. We know that sum of the eigenvalues is equal to the trace of matrix A, so 3 1 () . i i trace A Since we are given λ 1 =13.507, λ 2 =0.207, the third eigenvalue must be λ 3 = 0.714. b. We know that the determinant of a matrix is the product of eigenvalues. Therefore, 2 ) det( 3 1 i i A Note that we did not have to calculate the eigenvalues or the determinant explicitly using the data in the matrix A. You could have done the problem without having the entries of A! 4. a. We know that AB and BA will have the same non zero eigenvalues. Since BA will be 4×4 matrix, it will have 4 eigenvalues. Therefore, eigenvalues of BA must be 2, 4,
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EEL3105_MT2_Solution - /Fall2011 14 6 28 6 1.a...

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