Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# Following the solution to problem 30 with 3s instead

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Unformatted text preview: 4  4  1 D 64 16 D 48. Four zeros in the same row guarantee det D 0. A D I has 12 zeros (maximum with det ¤ 0). (a) If a11 D a22 D a33 D 0 then 4 terms are sure zeros (b) 15 terms must be zero. 2 det A D 3 4 5 6 54 Solutions to Exercises 7 5Š=2 D 60 permutation matrices have det D C1. Move row 5 of I to the top; starting 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 from .5; 1; 2; 3; 4/ elimination will do four row exchanges. Some term a1˛ a2ˇ    an! in the big formula is not zero! Move rows 1, 2, . . ., n into rows ˛ , ˇ , . . ., ! . Then these nonzero a’s will be on the main diagonal. To get C1 for the even permutations, the matrix needs an even number of 1’s. To get C1 for the odd P ’s, the matrix needs an odd number of 1’s. So all six terms D C1 in the big formula and det D 6 are impossible: max.det/ D 4. The 4Š=2 D 12 even permutations are .1; 2; 3; 4/; .2; 1; 4; 3/; .3; 1; 4; 2/; .4; 3; 2; 1/, and 8 P’s with one number in place and even permutation of the other three numbers. det.I C Peven ) D 16 or 4 or 0 (16 comes from I C I ). &quot; #   0 42 35 det B D 1.0/ C 2.42/ C 3. 35/ D 21. d b 0 21 14 . CD .DD c a Puzzle: det D D 441 D . 21/2 . Why? 3 6 3 &quot; # &quot; # 321 400 T C D 2 4 2 and AC D 0 4 0 . Therefore A 1 D 1 C T D C T = det A. 4 123 004 (a) C1 D 0, C2 D 1, C3 D 0, C4 D 1 (b) Cn D Cn 2 by cofactors of row 1 then cofactors of column 1. Therefore C10 D C8 D C6 D C4 D C2 D 1. We must choose 1’s from column 2 then column 1, column 4 then column 3,and so on. Therefore n must be even to have det An ¤ 0. The number of row exchanges is n=2 so Cn D . 1/n=2 . The 1; 1 cofactor of the n by n matrix is En 1 . The 1; 2 cofactor has a single 1 in its ﬁrst column, with cofactor En 2 : sign gives En 2 . So En D En 1 En 2 . Then E1 to E6 is 1, 0, 1, 1, 0, 1 and this cycle of six will repeat: E100 D E4 D 1. The 1; 1 cofactor of the n by n matrix is Fn 1 . The 1; 2 cofactor has a 1 in column 1, with cofactor Fn 2 . Multiply by . 1/1C2 and also . 1/ from the 1; 2 entry to ﬁnd Fn D Fn 1 C Fn 2 (so these determinants are Fibonacci numbers). &quot; # &quot; #   1 1 1 1 1 1 1 2 1 C det 1 2 jB4 j D 2 det D 2jB3 j det D 1 2 1 2 1 1 2jB3 j jB2 j. jB3 j and jB2 j are cofactors of row 4 of B4 . Rule 3 (linearity in row 1) gives jBn j D jAn j jAn 1 j D .n C 1/ n D 1. Since x , x 2 , x 3 are all in the same row, they are never multiplied in det V4 . The determinant is zero at x D a or b or c , so det V has factors .x a/.x b /.x c /. Multiply by the cofactor V3 . The Vandermonde matrix Vij D .xi /j 1 is for ﬁtting a polynomial p.x / D b at the points xi . It has det V D product of all xk xm for k &gt; m. G2 D 1, G3 D 2, G4 D 3, and Gn D . 1/n 1 .n 1/ D (product of the ’s ). S1 D 3; S2 D 8; S3 D 21. The rule looks like every second number in Fibonacci’s sequence : : : 3; 5; 8; 13; 21; 34; 55; : : : so the guess is S4 D 55. Following the solution to Problem 30 with 3’s instead of 2’s conﬁrms S4 D 81 C 1 9 9 9 D 55. Problem 33 directly proves Sn D F2nC...
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## This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

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