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# Discovering Algebra: An Investigative Approach 2nd Edition - Solutions by Chapter

## Full solutions for Discovering Algebra: An Investigative Approach | 2nd Edition

ISBN: 9781559537636

Discovering Algebra: An Investigative Approach | 2nd Edition - Solutions by Chapter

Solutions by Chapter
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##### ISBN: 9781559537636

This expansive textbook survival guide covers the following chapters: 90. This textbook survival guide was created for the textbook: Discovering Algebra: An Investigative Approach, edition: 2. Since problems from 90 chapters in Discovering Algebra: An Investigative Approach have been answered, more than 4324 students have viewed full step-by-step answer. Discovering Algebra: An Investigative Approach was written by and is associated to the ISBN: 9781559537636. The full step-by-step solution to problem in Discovering Algebra: An Investigative Approach were answered by , our top Math solution expert on 03/13/18, 07:06PM.

Key Math Terms and definitions covered in this textbook
• Basis for V.

Independent vectors VI, ... , v d whose linear combinations give each vector in V as v = CIVI + ... + CdVd. V has many bases, each basis gives unique c's. A vector space has many bases!

• 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.

• Elimination matrix = Elementary matrix Eij.

The identity matrix with an extra -eij in the i, j entry (i #- j). Then Eij A subtracts eij times row j of A from row i.

• Elimination.

A sequence of row operations that reduces A to an upper triangular U or to the reduced form R = rref(A). Then A = LU with multipliers eO in L, or P A = L U with row exchanges in P, or E A = R with an invertible E.

• 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.

• Linearly dependent VI, ... , Vn.

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

• 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.

• 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).

• Pivot columns of A.

Columns that contain pivots after row reduction. These are not combinations of earlier columns. The pivot columns are a basis for the column space.

• Pivot.

The diagonal entry (first nonzero) at the time when a row is used in elimination.

• 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.

• Reflection matrix (Householder) Q = I -2uuT.

Unit vector u is reflected to Qu = -u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q-1 = Q.

• Rotation matrix

R = [~ CS ] rotates the plane by () and R- 1 = RT rotates back by -(). Eigenvalues are eiO and e-iO , eigenvectors are (1, ±i). c, s = cos (), sin ().

• Row picture of Ax = b.

Each equation gives a plane in Rn; the planes intersect at x.

• Schur complement S, D - C A -} B.

Appears in block elimination on [~ g ].

• Triangle inequality II u + v II < II u II + II v II.

For matrix norms II A + B II < II A II + II B II·

• Unitary matrix UH = U T = U-I.

Orthonormal columns (complex analog of Q).

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