Exam 2 Solution

# If there is an invertible matrix such that 6 note dont

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Unformatted text preview: ix such that . 6. (Note: don’t include the factorization here). The Diagonalization Theorem states that an is diagonalizable if and only if has linearly independent eigenvectors. x matrix Part II. [4 pts each] Short answer / multiple choice. Only the answer(s) will be graded. Circle your answer(s). Problems 1 and 2 concern row equivalent matrices 2 1 3 5 4 2 6 10 1 4 6 5 3 3 7 10 5 7 15 20 1. Find (i) a basis for the column space of 2 1 , 3 5 (i) 2 (ii) 1 2 4 7 , 6 13 5 10 4 1 and below: 2 7 , 13 10 2 0 0 0 4 0 0 0 , 3 5 2, 0 0 3 4, 0 13 The null space of A is a k-dimensional subspace of 0 0 0 . 6 020 1,0,0 103 3. Consider B = a basis for . If x 2 5 14 4. For 3 9 4 9 6 9 0 0 9 5 9 7 352 134 005 000 and (ii) a basis for the row space of 2. What values of k and p make the following statement true? 3, 1 3 0 0 2 0 , given that det A = 12, find c. 0 0 5 1 , find x. 3 0 5 . 15 , the eigenvalues of 23 5. For are: a) 1, 3 b) 1+i, 1-i c) 2, 3 d) 2+i, 2-i e) 2+i, 2 f) 2+3i, 2-3i g) 5, -2 h) 1,-2...
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## This note was uploaded on 01/17/2014 for the course ESE 309 taught by Professor Staff during the Fall '08 term at Washington University in St. Louis.

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