- 8.1.1: Exer. 116: Solve .
- 8.1.2: Exer. 116: Solve .
- 8.1.3: Exer. 116: Solve .
- 8.1.4: Exer. 116: Solve .
- 8.1.5: Exer. 116: Solve .
- 8.1.6: Exer. 116: Solve .
- 8.1.7: Exer. 116: Solve .
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- 8.1.9: Exer. 116: Solve .
- 8.1.10: Exer. 116: Solve .
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- 8.1.14: Exer. 116: Solve .
- 8.1.15: Exer. 116: Solve .
- 8.1.16: Exer. 116: Solve .
- 8.1.17: To find the distance between two points A and B that lie on opposit...
- 8.1.18: To determine the distance between two points A and B, a surveyor ch...
- 8.1.19: As shown in the figure on the next page, a cable car carries passen...
- 8.1.20: A straight road makes an angle of 15 with the horizontal. When the ...
- 8.1.21: The angles of elevation of a balloon from two points A and B on lev...
- 8.1.22: Shown in the figure is a solar panel 10 feet in width, which is to ...
- 8.1.23: A straight road makes an angle of 22 with the horizontal. From a ce...
- 8.1.24: A surveyor notes that the direction from point A to point B is S63W...
- 8.1.25: A forest ranger at an observation point A sights a fire in the dire...
- 8.1.26: The leaning tower of Pisa was originally perpendicular to the groun...
- 8.1.27: A cathedral is located on a hill, as shown in the figure. When the ...
- 8.1.28: A helicopter hovers at an altitude that is 1000 feet above a mounta...
- 8.1.29: The volume V of the right triangular prism shown in the figure is w...
- 8.1.30: Shown in the figure on the next page is a plan for the top of a win...
- 8.1.31: Computer software for surveyorsmakes use of coordinate systems to l...
Solutions for Chapter 8.1: The Law of Sines
Full solutions for Algebra and Trigonometry with Analytic Geometry | 12th Edition
Tv = Av + Vo = linear transformation plus shift.
Characteristic equation det(A - AI) = O.
The n roots are the eigenvalues of A.
Commuting matrices AB = BA.
If diagonalizable, they share n eigenvectors.
Cross product u xv in R3:
Vector perpendicular to u and v, length Ilullllvlll sin el = area of parallelogram, u x v = "determinant" of [i j k; UI U2 U3; VI V2 V3].
Eigenvalue A and eigenvector x.
Ax = AX with x#-O so det(A - AI) = o.
The nullspace N (A) and row space C (AT) are orthogonal complements in Rn(perpendicular from Ax = 0 with dimensions rand n - r). Applied to AT, the column space C(A) is the orthogonal complement of N(AT) in Rm.
Hilbert matrix hilb(n).
Entries HU = 1/(i + j -1) = Jd X i- 1 xj-1dx. Positive definite but extremely small Amin and large condition number: H is ill-conditioned.
Identity matrix I (or In).
Diagonal entries = 1, off-diagonal entries = 0.
A symmetric matrix with eigenvalues of both signs (+ and - ).
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.
Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.
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.
Nilpotent matrix N.
Some power of N is the zero matrix, N k = o. The only eigenvalue is A = 0 (repeated n times). Examples: triangular matrices with zero diagonal.
Particular solution x p.
Any solution to Ax = b; often x p has free variables = o.
Ps = pascal(n) = the symmetric matrix with binomial entries (i1~;2). Ps = PL Pu all contain Pascal's triangle with det = 1 (see Pascal in the index).
Pseudoinverse A+ (Moore-Penrose inverse).
The n by m matrix that "inverts" A from column space back to row space, with N(A+) = N(AT). A+ A and AA+ are the projection matrices onto the row space and column space. Rank(A +) = rank(A).
Reduced row echelon form R = rref(A).
Pivots = 1; zeros above and below pivots; the r nonzero rows of R give a basis for the row space of A.
Schur complement S, D - C A -} B.
Appears in block elimination on [~ g ].
Transpose matrix AT.
Entries AL = Ajj. AT is n by In, AT A is square, symmetric, positive semidefinite. The transposes of AB and A-I are BT AT and (AT)-I.
Vandermonde matrix V.
V c = b gives coefficients of p(x) = Co + ... + Cn_IXn- 1 with P(Xi) = bi. Vij = (Xi)j-I and det V = product of (Xk - Xi) for k > i.