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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 ndimensional 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 4dimensional 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 @r=@x @r=@y ˇ ˇ x=r
y=r ˇ ˇ cos
sin ˇ
ˇ
ˇDˇ
ˇDˇ
ˇ
From x; y to r; : ˇ
@=@x @=@y 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 12 (with columns a < b ) times
a 3 by 3 determinant from rows 35 (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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 Spring '12
 Minki
 Mass

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