Introduction to Linear algebra-Strang-Solutions-Manual_ver13

41 the cauchy binet formula gives the determinant of

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Unformatted text preview: T3 2T 3 a aa 0 0  det AT A D .kakkb kkc k/2 AT A D 4 bT 5 a b c D 4 0 bT b 0 5 has det A D ˙kakkb kkc k cT 0 0 cTc " # 100 The box has height 4 and volume D det 0 1 0 D 4. i  j D k and .k  w/ D 4. 234 20 The edges of the hypercube have length 21 22 23 24 25 The n-dimensional cube has 2n corners, n2n 1 edges and 2n .n 1/-dimensional faces. Coefﬁcients from .2 C x/ in Worked Example 2.4A. Cube from 2I has volume 2n . 1 1 The pyramid has volume 1 . The 4-dimensional pyramid has volume 24 (and nŠ in Rn ) 6 x D r cos ; y D r sin  give J D r . The columns are orthogonal and their lengths are 1 and r . ˇ ˇ ˇ sin ' cos   cos ' sin   sin ' sin  ˇ ˇ ˇ J D ˇ sin ' sin   cos ' sin   sin ' cos  ˇ D 2 sin ' . This Jacobian is needed ˇ ˇ cos '  sin '  for triple integrals inside spheres. ˇ ˇˇ ˇˇ ˇ ˇ @[email protected] @[email protected] ˇ ˇ x=r y=r ˇ ˇ cos  sin  ˇ ˇ ˇDˇ ˇDˇ ˇ From x; y to r;  : ˇ @[email protected] @[email protected] ˇ ˇ y =r 2 x=r 2 ˇ ˇ . sin  /=r .cos  /=r ˇ 1 1 . DD r Jacobian in 27 The triangle with corners .0; 0/; .6; 0/; .1; 4/ has area 24. Rotated by  D 60ı the area ˇ ˇ ˇ cos  sin  ˇ ˇ ˇD is unchanged. The determinant of the rotation matrix is J D ˇ sin  cos  ˇ ˇ ˇ p ˇ 1=2 3=2 ˇ ˇp ˇ D 1. ˇ 3=2 1=2 ˇ n 26 27 28 29 30 31 Base area 10, height 2, volume 20. " 2 1 32 The volume of the box is det 1 ˇ ˇ ˇ ˇ ˇ u1 u2 u3 ˇ ˇ v2 v3 ˇ ˇ ˇ ˇ 33 ˇ v1 v2 v3 ˇ D u1 ˇ ˇ w2 w3 ˇ ˇw w w ˇ 1 2 3 # 40 3 0 D 20. 22 ˇ ˇ ˇ ˇ v1 v3 ˇ ˇ v1 ˇ ˇ u2 ˇ ˇ w 1 w 3 ˇ C u3 ˇ w 1 ˇ v2 ˇ ˇ. This is u  .v  w/. w2 ˇ 34 .w  u/  v D .v  w/  u D .u  v/  w W Even permutation of .u; v; w/ keeps the same determinant. Odd permutations reverse the sign. Solutions to Exercises 58 35 S D .2; 1; 1/, area kPQ  PS k D k. 2 ; 2 ; 1/k D 3. The other four corners can be .0; 0; 0/, .0; 0; 2/, .1; 2; 2/, .1; 1; 0/. The volume of the tilted box is j det j D 1. 2 3 xyz 36 If .1; 1; 0/, .1; 2; 1/, .x; y; z/ are in a plane the volume is det 4 1 1 0 5 D x y C z D 0. The “box” with those edges is ﬂattened to zero height. 121 " # xyz 37 det 2 3 1 D 7x 5y C z will be zero when .x; y; z/ is a combination of .2; 3; 1/ 123 and .1; 2; 3/. The plane containing those two vectors has equation 7x 5y C z D 0. 38 Doubling each row multiplies the volume by 2n . Then 2 det A D det.2A/ only if n D 1. D .det A/I gives .det A/.det C / D .det A/n . Then det A D .det C /1=3 with n D 4. With det A 1 D 1= det A, construct A 1 using the cofactors. Invert to ﬁnd A. 39 AC T 40 The cofactor formula adds 1 by 1 determinants (which are just entries) times their co- factors of size n 1. Jacobi discovered that this formula can be generalized. For n D 5, Jacobi multiplied each 2 by 2 determinant from rows 1-2 (with columns a < b ) times a 3 by 3 determinant from rows 3-5 (using the remaining columns c < d < e ). The key question is C or sign (as for cofactors). The product is given a C sign when a, b , c , d , e is an even permutation of 1, 2, 3, 4, 5. This gives the...
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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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