# 16 the 1 1 cofactor of the n by n matrix is f n nul 1

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16 The 1; 1 cofactor of the n by n matrix is F n NUL 1 . The 1; 2 cofactor has a 1 in column 1, with cofactor F n NUL 2 . Multiply by . NUL 1/ 1 C 2 and also . NUL 1/ from the 1; 2 entry to find F n D F n NUL 1 C F n NUL 2 (so these determinants are Fibonacci numbers). 17 j B 4 j D 2 det " 1 NUL 1 NUL 1 2 NUL 1 NUL 1 2 # C det " 1 NUL 1 NUL 1 2 NUL 1 NUL 1 # D 2 j B 3 j NUL det ± 1 NUL 1 NUL 1 2 ² D 2 j B 3 j NUL j B 2 j . j B 3 j and NULj B 2 j are cofactors of row 4 of B 4 . 18 Rule 3 (linearity in row 1) gives j B n j D j A n j NUL j A n NUL 1 j D .n C 1/ NUL n D 1 . 19 Since x , x 2 , x 3 are all in the same row, they are never multiplied in det V 4 . The deter- minant is zero at x D a or b or c , so det V has factors .x NUL a/.x NUL b/.x NUL c/ . Multiply by the cofactor V 3 . The Vandermonde matrix V ij D .x i / j NUL 1 is for fitting a polynomial p. x / D b at the points x i . It has det V D product of all x k NUL x m for k > m . 20 G 2 D NUL 1 , G 3 D 2 , G 4 D NUL 3 , and G n D . NUL 1/ n NUL 1 .n NUL 1/ D (product of the ´ ’s ). 21 S 1 D 3; S 2 D 8; S 3 D 21 . The rule looks like every second number in Fibonacci’s sequence : : : 3; 5; 8; 13; 21; 34; 55; : : : so the guess is S 4 D 55 . Following the solution to Problem 30 with 3’s instead of 2’s confirms S 4 D 81 C 1 NUL 9 NUL 9 NUL 9 D 55 . Problem 33 directly proves S n D F 2n C 2 . 22 Changing 3 to 2 in the corner reduces the determinant F 2n C 2 by 1 times the cofactor of that corner entry. This cofactor is the determinant of S n NUL 1 (one size smaller) which is F 2n . Therefore changing 3 to 2 changes the determinant to F 2n C 2 NUL F 2n which is F 2n C 1 .
Solutions to Exercises 55 23 (a) If we choose an entry from B we must choose an entry from the zero block; re- sult zero. This leaves entries from A times entries from D leading to . det A/. det D/ (b) and (c) Take A D ± 1 0 0 0 ² , B D ± 0 0 1 0 ² , C D ± 0 1 0 0 ² , D D ± 0 0 0 1 ² . See # 25 . 24 (a) All L ’s have det D 1 I det U k D det A k D 2; 6; NUL 6 for k D 1; 2; 3 (b) Pivots 2; 3 2 ; NUL 1 3 . 25 Problem 23 gives det ± I 0 NUL CA NUL 1 I ² D 1 and det ± A B C D ² D j A j times j D NUL CA NUL 1 B j which is j AD NUL ACA NUL 1 B j . If AC D CA this is j AD NUL CAA NUL 1 B j D det .AD NUL CB/ . 26 If A is a row and B is a column then det M D det AB D dot product of A and B . If A is a column and B is a row then AB has rank 1 and det M D det AB D 0 (unless m D n D 1 ). This block matrix is invertible when AB is invertible which certainly requires m ² n . 27 (a) det A D a 11 C 11 C ´ ´ ´ C a 1n C 1n . Derivative with respect to a 11 D cofactor C 11 . 28 Row 1 NUL 2 row 2 C row 3 D 0 so this matrix is singular. 29 There are five nonzero products, all 1’s with a plus or minus sign. Here are the (row, column) numbers and the signs: C .1; 1/.2; 2/.3; 3/.4; 4/ C .1; 2/.2; 1/.3; 4/.4; 3/ NUL .1; 2/.2; 1/.3; 3/.4; 4/ NUL .1; 1/.2; 2/.3; 4/.4; 3/ NUL .1; 1/.2; 3/.3; 2/.4; 4/ . Total NUL 1 . 30 The 5 products in solution 29 change to 16 C 1 NUL 4 NUL 4 NUL 4 since A has 2’s and -1’s: .2/.2/.2/.2/ C . NUL 1/. NUL 1/. NUL 1/. NUL 1/ NUL . NUL 1/. NUL 1/.2/.2/ NUL .2/.2/. NUL 1/. NUL 1/ NUL .2/. NUL 1/. NUL 1/.2/: 31 det P D NUL 1 because the cofactor of P 14 D 1 in row one has sign . NUL 1/ 1 C 4 . The big formula for det P