quiz4_sol

quiz4_sol - Math 33A, Section 3, Fall 2009 Quiz #4...

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Unformatted text preview: Math 33A, Section 3, Fall 2009 Quiz #4 Solutions 1. 23 = 2 · 6 − 3 · 4 = 12 − 12 = 0 46 2. −9 4 = (−9) · 1 − 4 · (−2) = −9 + 8 = −1 −2 1 3. There is only one non-zero pattern, and it has 3 inversions. 002 0 3 0 = −2 · 3 · 5 = −30 500 4. We subtract 2 times row I from row II, and 3 times row I from row III. 111 111 2 3 4 = 0 1 2 =1·1·3=3 336 003 5. We first factor out 1 and 1 from rows I and III respectively, and then realize that the 3 5 obtained determinant has two equal rows. 2 3 7 3 5 789 = 27 3 55 1 3 2 7 15 ·1· 7 8 9 =0 5 2 7 15 6. If we subtract row I from row III, we get precisely row II. 3 2 5 6 2 3 5 7 4 1 5 8 1 32 4 23 = 5 23 9 67 4 1 1 8 1 4 =0 4 9 7. There is only one non-zero pattern, and it has 6 inversions. 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 =1·1·1·1·1·1=1 0 0 1 8. The matrix is lower-triangular. 1 2 3 4 5 0 −1 −2 −3 −4 00 00 10 2 −1 3 −2 0 0 0 = 1 · (−1) · 1 · (−1) · 1 = 1 0 1 9. There is only one non-zero pattern, and it has 5 inversions. 0 8 1 0 0 0 3 0 0 0 5 8 9 4 7 9 6 4 0 1 2 1 3 = −2 · 3 · 1 · 4 · 1 = −24 0 0 10. The matrix is block-upper-triangular and each of the blocks is lower-triangular. 10 5 −2 46 00 00 00 0 5 −7 9 0 8 6 −3 100 300 24 5 7 = 5 −2 0 · 9 1 0 = (1 · (−2) · 2) · (3 · 1 · (−1)) = 12 03 0 0 462 −9 8 −1 09 1 0 0 −9 8 −1 V. K. 2 ...
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This note was uploaded on 05/29/2011 for the course MATH 33a taught by Professor Lee during the Spring '08 term at UCLA.

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