 7.6.1: . Complete the table by filling in the missing values for the side,...
 7.6.2: Solve each equation for x.a.  x  = 6 b. x2 = 36 c.  x  = 3.8 d....
 7.6.3: Solve each equation, if possible. a. 4.7 =  x  2.8 b. 41 = x2 2.8...
 7.6.4: Solve each equation for x. Use a calculator graph to check your ans...
 7.6.5: For what values of x is  x  x2? To check your answer, graph Y1 = ...
 7.6.6: For what values of y does the equation y = x2 have a. No real solut...
 7.6.7: Graph the function What other equation produces the same graph?
 7.6.8: Look at the table of squares in the Investigation Graphing a Parabo...
 7.6.9: MiniInvestigation Find the sum of each set of numbers in 9ac. a. t...
 7.6.10: Write an equation for the function represented in each table. Use y...
 7.6.11: This 4by4 grid contains squares of different sizes. a. How many o...
 7.6.12: Explain why the equation x2 = 4 has no solutions.
 7.6.13: The table shows exponential data. xy 0 4 126.5625 3 168.75 1 1000 a...
 7.6.14: Use properties of exponents to find an equivalent expression in the...
 7.6.15: Graph the functions f (x) = 3x 5 and g(x) =  x 3 . What do the tw...
Solutions for Chapter 7.6: Squares, Squaring, and Parabolas
Full solutions for Discovering Algebra: An Investigative Approach  2nd Edition
ISBN: 9781559537636
Solutions for Chapter 7.6: Squares, Squaring, and Parabolas
Get Full SolutionsThis textbook survival guide was created for the textbook: Discovering Algebra: An Investigative Approach, edition: 2. This expansive textbook survival guide covers the following chapters and their solutions. Discovering Algebra: An Investigative Approach was written by Patricia and is associated to the ISBN: 9781559537636. Since 15 problems in chapter 7.6: Squares, Squaring, and Parabolas have been answered, more than 2877 students have viewed full stepbystep solutions from this chapter. Chapter 7.6: Squares, Squaring, and Parabolas includes 15 full stepbystep solutions.

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

Column picture of Ax = b.
The vector b becomes a combination of the columns of A. The system is solvable only when b is in the column space C (A).

Column space C (A) =
space of all combinations of the columns of A.

Complex conjugate
z = a  ib for any complex number z = a + ib. Then zz = Iz12.

Diagonalization
A = S1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k SI.

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

Ellipse (or ellipsoid) x T Ax = 1.
A must be positive definite; the axes of the ellipse are eigenvectors of A, with lengths 1/.JI. (For IIx II = 1 the vectors y = Ax lie on the ellipse IIA1 yll2 = Y T(AAT)1 Y = 1 displayed by eigshow; axis lengths ad

Fourier matrix F.
Entries Fjk = e21Cijk/n give orthogonal columns FT F = nI. Then y = Fe is the (inverse) Discrete Fourier Transform Y j = L cke21Cijk/n.

Free variable Xi.
Column i has no pivot in elimination. We can give the n  r free variables any values, then Ax = b determines the r pivot variables (if solvable!).

Full column rank r = n.
Independent columns, N(A) = {O}, no free variables.

Fundamental Theorem.
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.

Hermitian matrix A H = AT = A.
Complex analog a j i = aU of a symmetric matrix.

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

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.

Orthogonal subspaces.
Every v in V is orthogonal to every w in W.

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 •

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.

Spanning set.
Combinations of VI, ... ,Vm fill the space. The columns of A span C (A)!

Spectral Theorem A = QAQT.
Real symmetric A has real A'S and orthonormal q's.

Vector space V.
Set of vectors such that all combinations cv + d w remain within V. Eight required rules are given in Section 3.1 for scalars c, d and vectors v, w.
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