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# Solutions for Chapter 2.5: The Point Slope-Form of the Equation of a Line

## Full solutions for Intermediate Algebra for College Students | 6th Edition

ISBN: 9780321758934

Solutions for Chapter 2.5: The Point Slope-Form of the Equation of a Line

Solutions for Chapter 2.5
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##### ISBN: 9780321758934

This textbook survival guide was created for the textbook: Intermediate Algebra for College Students, edition: 6. Chapter 2.5: The Point Slope-Form of the Equation of a Line includes 100 full step-by-step solutions. Intermediate Algebra for College Students was written by and is associated to the ISBN: 9780321758934. This expansive textbook survival guide covers the following chapters and their solutions. Since 100 problems in chapter 2.5: The Point Slope-Form of the Equation of a Line have been answered, more than 36760 students have viewed full step-by-step solutions from this chapter.

Key Math Terms and definitions covered in this textbook
• Affine transformation

Tv = Av + Vo = linear transformation plus shift.

• Cayley-Hamilton Theorem.

peA) = det(A - AI) has peA) = zero matrix.

• Column space C (A) =

space of all combinations of the columns of A.

• Companion matrix.

Put CI, ... ,Cn in row n and put n - 1 ones just above the main diagonal. Then det(A - AI) = ±(CI + c2A + C3A 2 + .•. + cnA n-l - An).

• Condition number

cond(A) = c(A) = IIAIlIIA-III = amaxlamin. In Ax = b, the relative change Ilox III Ilx II is less than cond(A) times the relative change Ilob III lib II· Condition numbers measure the sensitivity of the output to change in the input.

• Diagonalization

A = S-1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k S-I.

• 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.

• Hypercube matrix pl.

Row n + 1 counts corners, edges, faces, ... of a cube in Rn.

• 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.

• Inverse matrix A-I.

Square matrix with A-I A = I and AA-l = I. No inverse if det A = 0 and rank(A) < n and Ax = 0 for a nonzero vector x. The inverses of AB and AT are B-1 A-I and (A-I)T. Cofactor formula (A-l)ij = Cji! detA.

• Linear combination cv + d w or L C jV j.

• Linearly dependent VI, ... , Vn.

A combination other than all Ci = 0 gives L Ci Vi = O.

• Normal equation AT Ax = ATb.

Gives the least squares solution to Ax = b if A has full rank n (independent columns). The equation says that (columns of A)·(b - Ax) = o.

• Saddle point of I(x}, ... ,xn ).

A point where the first derivatives of I are zero and the second derivative matrix (a2 II aXi ax j = Hessian matrix) is indefinite.

• Singular matrix A.

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

• Solvable system Ax = b.

The right side b is in the column space of A.

• Special solutions to As = O.

One free variable is Si = 1, other free variables = o.

• Standard basis for Rn.

Columns of n by n identity matrix (written i ,j ,k in R3).

• Symmetric factorizations A = LDLT and A = QAQT.

Signs in A = signs in D.

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