- 9.5.1: is a triangular display of the binomial coefficients.
- 9.5.2: a b = n 0 b = and . an1 a b =
- 9.5.3: True or False anjb = j!1n - j2! n!
- 9.5.4: The can be used to expand expressions like 12x + 326 .
- 9.5.5: a 5 3 b
- 9.5.6: a 7 3 b
- 9.5.7: a 7 5 b
- 9.5.8: a 9 7 b
- 9.5.9: a 50 49 b
- 9.5.10: a 100 98 b
- 9.5.11: a 1000 1000 b
- 9.5.12: a 1000 0 b
- 9.5.13: a 55 23 b
- 9.5.14: a 60 20b
- 9.5.15: a 47 25 b
- 9.5.16: a 37 19 b
- 9.5.17: 1x + 125
- 9.5.18: 1x - 125
- 9.5.19: 1x - 226
- 9.5.20: 1x + 325
- 9.5.21: 13x + 124
- 9.5.22: 12x + 325
- 9.5.23: 1x2 + y2 2 5
- 9.5.24: 1x2 - y2 2 6
- 9.5.25: A 1x + 22B 6
- 9.5.26: A 1x - 23B 4
- 9.5.27: 1ax + by25
- 9.5.28: 1ax - by24
- 9.5.29: The coefficient of in the expansion of 1x + 3210
- 9.5.30: The coefficient of in the expansion of 1x - 3210
- 9.5.31: The coefficient of in the expansion of 12x - 1212
- 9.5.32: The coefficient of in the expansion of 12x + 1212
- 9.5.33: The coefficient of in the expansion of 12x + 32
- 9.5.34: The coefficient of in the expansion of 12x - 329
- 9.5.35: The fifth term in the expansion of 1x + 327
- 9.5.36: The third term in the expansion of 1x - 327
- 9.5.37: The third term in the expansion of 13x - 229
- 9.5.38: The sixth term in the expansion of 13x + 228
- 9.5.39: The coefficient of in the expansion of ax2 + 1 x b 12
- 9.5.40: The coefficient of in the expansion of x - 1 x2 9
- 9.5.41: The coefficient of in the expansion of ax - 2 1x b 10
- 9.5.42: The coefficient of in the expansion of a 1x + 3 1x b 8
- 9.5.43: Use the Binomial Theorem to find the numerical value of correct to ...
- 9.5.44: Use the Binomial Theorem to find the numerical value of correct to ...
- 9.5.45: Show that and ann a b = 1.
- 9.5.46: Show that if n and j are integers with then anjb = a nn - jb Conclu...
- 9.5.47: f n is a positive integer, show that [Hint: 2 now use the Binomial ...
- 9.5.48: If n is a positive integer, show that an0b - an1b + an2b - + 1-12n ...
- 9.5.49: a 5 0 b a 1 4 b 5 + a 5 1 b a 1 4 b 4 a 3 4 b + a 5 2 b a 1 4 b 3 a...
- 9.5.50: Stirlings Formula An approximation for n!, when n is large, is give...
Solutions for Chapter 9.5: The Binomial Theorem
Full solutions for College Algebra | 9th Edition
Basis for V.
Independent vectors VI, ... , v d whose linear combinations give each vector in V as v = CIVI + ... + CdVd. V has many bases, each basis gives unique c's. A vector space has many bases!
A matrix can be partitioned into matrix blocks, by cuts between rows and/or between columns. Block multiplication ofAB is allowed if the block shapes permit.
peA) = det(A - AI) has peA) = zero matrix.
Characteristic equation det(A - AI) = O.
The n roots are the eigenvalues of A.
z = a - ib for any complex number z = a + ib. Then zz = Iz12.
Conjugate Gradient Method.
A sequence of steps (end of Chapter 9) to solve positive definite Ax = b by minimizing !x T Ax - x Tb over growing Krylov subspaces.
Determinant IAI = det(A).
Defined by det I = 1, sign reversal for row exchange, and linearity in each row. Then IAI = 0 when A is singular. Also IABI = IAIIBI and
Diagonalizable matrix A.
Must have n independent eigenvectors (in the columns of S; automatic with n different eigenvalues). Then S-I AS = A = eigenvalue matrix.
Invert A by row operations on [A I] to reach [I A-I].
Hankel matrix H.
Constant along each antidiagonal; hij depends on i + j.
Hessenberg matrix H.
Triangular matrix with one extra nonzero adjacent diagonal.
Identity matrix I (or In).
Diagonal entries = 1, off-diagonal entries = 0.
Independent vectors VI, .. " vk.
No combination cl VI + ... + qVk = zero vector unless all ci = O. If the v's are the columns of A, the only solution to Ax = 0 is x = o.
Markov matrix M.
All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.
Orthogonal matrix Q.
Square matrix with orthonormal columns, so QT = Q-l. Preserves length and angles, IIQxll = IIxll and (QX)T(Qy) = xTy. AlllAI = 1, with orthogonal eigenvectors. Examples: Rotation, reflection, permutation.
Plane (or hyperplane) in Rn.
Vectors x with aT x = O. Plane is perpendicular to a =1= O.
Projection p = a(aTblaTa) onto the line through a.
P = aaT laTa has rank l.
Rayleigh quotient q (x) = X T Ax I x T x for symmetric A: Amin < q (x) < Amax.
Those extremes are reached at the eigenvectors x for Amin(A) and Amax(A).
Special solutions to As = O.
One free variable is Si = 1, other free variables = o.
Tridiagonal matrix T: tij = 0 if Ii - j I > 1.
T- 1 has rank 1 above and below diagonal.