Solutions for Chapter 18: Complex Numbers and Polar Coordinates
Full solutions for Trigonometry  7th Edition
ISBN: 9781111826857
Solutions for Chapter 18: Complex Numbers and Polar Coordinates
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Associative Law (AB)C = A(BC).
Parentheses can be removed to leave ABC.

Augmented matrix [A b].
Ax = b is solvable when b is in the column space of A; then [A b] has the same rank as A. Elimination on [A b] keeps equations correct.

Characteristic equation det(A  AI) = O.
The n roots are the eigenvalues of A.

Diagonalizable matrix A.
Must have n independent eigenvectors (in the columns of S; automatic with n different eigenvalues). Then SI AS = A = eigenvalue matrix.

Factorization
A = L U. If elimination takes A to U without row exchanges, then the lower triangular L with multipliers eij (and eii = 1) brings U back to A.

GaussJordan method.
Invert A by row operations on [A I] to reach [I AI].

Hessenberg matrix H.
Triangular matrix with one extra nonzero adjacent diagonal.

lAII = l/lAI and IATI = IAI.
The big formula for det(A) has a sum of n! terms, the cofactor formula uses determinants of size n  1, volume of box = I det( A) I.

Network.
A directed graph that has constants Cl, ... , Cm associated with the edges.

Norm
IIA II. The ".e 2 norm" of A is the maximum ratio II Ax II/l1x II = O"max· Then II Ax II < IIAllllxll and IIABII < IIAIIIIBII and IIA + BII < IIAII + IIBII. Frobenius norm IIAII} = L La~. The.e 1 and.e oo norms are largest column and row sums of laij I.

Normal matrix.
If N NT = NT N, then N has orthonormal (complex) eigenvectors.

Nullspace matrix N.
The columns of N are the n  r special solutions to As = O.

Orthonormal vectors q 1 , ... , q n·
Dot products are q T q j = 0 if i =1= j and q T q i = 1. The matrix Q with these orthonormal columns has Q T Q = I. If m = n then Q T = Q 1 and q 1 ' ... , q n is an orthonormal basis for Rn : every v = L (v T q j )q j •

Particular solution x p.
Any solution to Ax = b; often x p has free variables = o.

Pascal matrix
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).

Polar decomposition A = Q H.
Orthogonal Q times positive (semi)definite H.

Reflection matrix (Householder) Q = I 2uuT.
Unit vector u is reflected to Qu = u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q1 = Q.

Right inverse A+.
If A has full row rank m, then A+ = AT(AAT)l has AA+ = 1m.

Similar matrices A and B.
Every B = MI AM has the same eigenvalues as A.

Singular matrix A.
A square matrix that has no inverse: det(A) = o.