- 9.9.11: In Exercises 712, identify the center and radius of the circle.
- 9.9.12: In Exercises 712, identify the center and radius of the circle.
Solutions for Chapter 9: Circles and Parabolas
Full solutions for Precalculus With Limits A Graphing Approach | 5th Edition
Adjacency matrix of a graph.
Square matrix with aij = 1 when there is an edge from node i to node j; otherwise aij = O. A = AT when edges go both ways (undirected). Adjacency matrix of a graph. Square matrix with aij = 1 when there is an edge from node i to node j; otherwise aij = O. A = AT when edges go both ways (undirected).
Characteristic equation det(A - AI) = O.
The n roots are the eigenvalues of A.
z = a - ib for any complex number z = a + ib. Then zz = Iz12.
Four Fundamental Subspaces C (A), N (A), C (AT), N (AT).
Use AT for complex A.
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!).
Set of n nodes connected pairwise by m edges. A complete graph has all n(n - 1)/2 edges between nodes. A tree has only n - 1 edges and no closed loops.
A sequence of steps intended to approach the desired solution.
Linear combination cv + d w or L C jV j.
Vector addition and scalar multiplication.
Nullspace matrix N.
The columns of N are the n - r special solutions to As = O.
Orthogonal matrix Q.
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.
Every v in V is orthogonal to every w in W.
The diagonal entry (first nonzero) at the time when a row is used in elimination.
Positive definite matrix A.
Symmetric matrix with positive eigenvalues and positive pivots. Definition: x T Ax > 0 unless x = O. Then A = LDLT with diag(D» O.
Projection p = a(aTblaTa) onto the line through a.
P = aaT laTa has rank l.
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.
Right inverse A+.
If A has full row rank m, then A+ = AT(AAT)-l has AA+ = 1m.
Solvable system Ax = b.
The right side b is in the column space of A.
Symmetric factorizations A = LDLT and A = QAQT.
Signs in A = signs in D.
Constant down each diagonal = time-invariant (shift-invariant) filter.
Transpose matrix AT.
Entries AL = Ajj. AT is n by In, AT A is square, symmetric, positive semidefinite. The transposes of AB and A-I are BT AT and (AT)-I.