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Solutions for Chapter 10.1  10.4: MIDCHAPTER CHECK POINT
Full solutions for Introductory & Intermediate Algebra for College Students  4th Edition
ISBN: 9780321758941
Solutions for Chapter 10.1  10.4: MIDCHAPTER CHECK POINT
Get Full SolutionsThis textbook survival guide was created for the textbook: Introductory & Intermediate Algebra for College Students, edition: 4. Chapter 10.1  10.4: MIDCHAPTER CHECK POINT includes 25 full stepbystep solutions. Since 25 problems in chapter 10.1  10.4: MIDCHAPTER CHECK POINT have been answered, more than 68301 students have viewed full stepbystep solutions from this chapter. Introductory & Intermediate Algebra for College Students was written by and is associated to the ISBN: 9780321758941. This expansive textbook survival guide covers the following chapters and their solutions.

Circulant matrix C.
Constant diagonals wrap around as in cyclic shift S. Every circulant is Col + CIS + ... + Cn_lSn  l . Cx = convolution c * x. Eigenvectors in F.

Cyclic shift
S. Permutation with S21 = 1, S32 = 1, ... , finally SIn = 1. Its eigenvalues are the nth roots e2lrik/n of 1; eigenvectors are columns of the Fourier matrix F.

Dot product = Inner product x T y = XI Y 1 + ... + Xn Yn.
Complex dot product is x T Y . Perpendicular vectors have x T y = O. (AB)ij = (row i of A)T(column j of B).

Eigenvalue A and eigenvector x.
Ax = AX with x#O so det(A  AI) = o.

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.

Fast Fourier Transform (FFT).
A factorization of the Fourier matrix Fn into e = log2 n matrices Si times a permutation. Each Si needs only nl2 multiplications, so Fnx and Fn1c can be computed with ne/2 multiplications. Revolutionary.

Fibonacci numbers
0,1,1,2,3,5, ... satisfy Fn = Fnl + Fn 2 = (A7 A~)I()q A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].

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

Hankel matrix H.
Constant along each antidiagonal; hij depends on i + j.

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.

Jordan form 1 = M 1 AM.
If A has s independent eigenvectors, its "generalized" eigenvector matrix M gives 1 = diag(lt, ... , 1s). The block his Akh +Nk where Nk has 1 's on diagonall. Each block has one eigenvalue Ak and one eigenvector.

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.

Linear combination cv + d w or L C jV j.
Vector addition and scalar multiplication.

Lucas numbers
Ln = 2,J, 3, 4, ... satisfy Ln = L n l +Ln 2 = A1 +A~, with AI, A2 = (1 ± /5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = O.

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

Projection matrix P onto subspace S.
Projection p = P b is the closest point to b in S, error e = b  Pb is perpendicularto S. p 2 = P = pT, eigenvalues are 1 or 0, eigenvectors are in S or S...L. If columns of A = basis for S then P = A (AT A) 1 AT.

Random matrix rand(n) or randn(n).
MATLAB creates a matrix with random entries, uniformly distributed on [0 1] for rand and standard normal distribution for randn.

Row space C (AT) = all combinations of rows of A.
Column vectors by convention.

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

Transpose matrix AT.
Entries AL = Ajj. AT is n by In, AT A is square, symmetric, positive semidefinite. The transposes of AB and AI are BT AT and (AT)I.