Intermediate Algebra 6th Edition - Solutions by Chapter

A matrix can be partitioned into matrix blocks, by cuts between rows and/or between columns. Block multiplication ofAB is allowed if the block shapes permit.

dij = 0 if i #- j. Block-diagonal: zero outside square blocks Du.

Ax = AX with x#-O so det(A - AI) = o.

has derivative AeAt; eAt u(O) solves u' = Au.

The i, j entry of AB is (row i of A)·(column j of B) = L aikbkj. By columns: Column j of AB = A times column j of B. By rows: row i of A multiplies B. Columns times rows: AB = sum of (column k)(row k). All these equivalent definitions come from the rule that A B times x equals A times B x .

The lowest degree polynomial with meA) = zero matrix. This is peA) = det(A - AI) if no eigenvalues are repeated; always meA) divides peA).

= All solutions to Ax = O. Dimension n - r = (# columns) - rank.

Square matrix with orthonormal columns, so QT = Q-l. Preserves length and angles, IIQxll = IIxll and (QX)T(Qy) = xTy. AlllAI = 1, with orthogonal eigenvectors. Examples: Rotation, reflection, permutation.

In each column, choose the largest available pivot to control roundoff; all multipliers have leij I < 1. See condition number.

There are n! orders of 1, ... , n. The n! P 's have the rows of I in those orders. P A puts the rows of A in the same order. P is even or odd (det P = 1 or -1) based on the number of row exchanges to reach I.

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

Vectors x with aT x = O. Plane is perpendicular to a =1= O.

Orthogonal Q times positive (semi)definite H.

Symmetric matrix with positive eigenvalues and positive pivots. Definition: x T Ax > 0 unless x = O. Then A = LDLT with diag(D» O.

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 ().

Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.

Every B = M-I AM has the same eigenvalues as A.

The minimum cost vector x * is found by moving from comer to lower cost comer along the edges of the feasible set (where the constraints Ax = b and x > 0 are satisfied). Minimum cost at a comer!

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

Signs in A = signs in D.