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v = Table[2k 1, {k, 1, 8}]
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3, 5, 7, 9, 11, 13, 15}
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norm = v.v 2 170
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12.9 Prove that the cross product of two vectors in 3 is orthogonal to each of the vectors that form it.
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Let u = (a, b, c) and v = (d, e, f) and compute w = u v. Then verify that w^u and w^v. Two vectors are orthogonal (^) if their dot product is 0. u = {u1, u2, u3}; v = {v1, v2, v3}; w = Cross[u, v]
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u3 v2 + u2 v3, u3 v1 u1 v3, u2 v1 + u1 v2}
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w.u // Expand 0 w.v // Expand 0
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12.10 It can be shown that the volume of a parallelepiped formed by u, v, and w is | u (v w) |. Compute the volume of the parallelepiped formed by i + 2 j 3 k, 2 i 5 j + k , and 3 i + j + 2 k. (The quantity u (v w) is called the scalar triple product.)
SOLUTION
u = {1, 2, 3}; v = {2, 5, 1}; w = {3, 1, 2}; volume = Abs[u.Cross[v, w]] 64
12.11 Let u = (u1, u2, u3), v = (v1, v2, v3), and w = (w1, w2, w3). Prove that the scalar triple product, u1 u2 u3 u (v w) = v1 v2 v3 . w1 w2 w3
SOLUTION
u = {u1, u2, u3}; v = {v1, v2, v3}; w = {w1, w2, w3}; matrix = {{u1, u2, u3}, {v1, v2, v3}, {w1, w2, w3}}; lhs = u.Cross[v, w] // Expand; rhs = Det[matrix] // Expand; lhs rhs True
12.12 Construct the Hilbert matrix of order 6, and compute its determinant and its inverse.
SOLUTION
hilbert = HilbertMatrix[6]; hilbert // MatrixForm 1 1 2 1 3 1 4 1 5 1 6 1 2 1 3 1 4 1 5 1 6 1 7 1 3 1 4 1 5 1 6 1 7 1 8 1 4 1 5 1 6 1 7 1 8 1 9 1 5 1 6 1 7 1 8 1 9 1 10 1 6 1 7 1 8 1 9 1 10 1 11
Det[hilbert] 1 186 313 420 339 200 000
MatrixForm[Inverse[hilbert], TableAlignments Right]
36 630 3 360 7 560 7 560 -2 772 630 14 700 -88 200 211 680 220 500 83 160 564 480 1 411 200 1 512 000 0 582 120 3 360 88 200 7 560 211 680 1 411 200 1 3 628 800 3 969 000 1 552 320 1 512 000 3 969 000 4 410 000 1 746 360 7 560 220 500 2 772 83 160 582 120 1 552 320 1 746 360 0 698 544
Linear Algebra
12.13 Construct a table that shows the determinant of the Hilbert matrices of orders 1 through 10.
SOLUTION
TableForm[Table[{k, Det[HilbertMatrix[k]] // N}, {k, 1, 10}], TableSpacing {1, 5}, TableHeadings {None, {"k", "determinant"}}]
k 1 2 3 4 5 6 7 8 9 10 determinant 1. 0.0833333 0.000462963 1.65344 10 7 3.7493 10 12 5.3673 10 18 4.8358 10 25 2.73705 10 33 9.72023 10 43 2.16418 10 53 Since the determinants are nonzero, each Hilbert matrix is invertible. The Hilbert matrix is a classic example of an ill-conditioned matrix.
1 10 12.14 Let M = 3 10 5 10
SOLUTION
2 10 3 10 4 10
7 10 4 . Compute lim M n . (M is a stochastic matrix.) n 10 1 10
1 2 7 m = 1 3 3 4 ; 10 5 4 1 Limit[MatrixPower[m, n], n // MatrixForm 47 150 47 150 47 150 23 19 75 50 23 19 75 50 23 19 75 50
1 2 1 2 3 1 2 2 0 2 0 1 2 3 1 and f ( x ) = x 5 + 2 x 4 x 3 + x 2 3 x + 2. Compute f(A). 12.15 Let A = 1 1 1 2 3 1 1 2 2 2
SOLUTION
1 2 1 2 3 2 1 2 2 0 a = 0 1 2 3 1 ; 1 1 1 2 3 2 2 1 1 2
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