- 1.1.1: Use back substitution to solve each of the following systems of equ...
- 1.1.2: Write out the coefficient matrix for each of the systems in Exercis...
- 1.1.3: In each of the following systems, interpret each equation as a line...
- 1.1.4: Write an augmented matrix for each of the systems in Exercise 3.
- 1.1.5: Write out the system of equations that corresponds to each of the f...
- 1.1.6: Solve each of the following systems. (a) x1 2x2 = 5 3x1 + x2 = 1 (b...
- 1.1.7: The two systems 2x1 + x2 = 3 4x1 + 3x2 = 5 and 2x1 + x2 = 1 4x1 + 3...
- 1.1.8: Solve the two systems x1 + 2x2 2x3 = 1 2x1 + 5x2 + x3 = 9 x1 + 3x2 ...
- 1.1.9: Given a system of the form m1x1 + x2 = b1 m2x1 + x2 = b2 where m1, ...
- 1.1.10: . Consider a system of the form a11x1 + a12x2 = 0 a21x1 + a22x2 = 0...
- 1.1.11: Give a geometrical interpretation of a linear equation in three unk...
Solutions for Chapter 1.1: Systems of Linear Equations
Full solutions for Linear Algebra with Applications | 9th 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).
Circulant matrix C.
Constant diagonals wrap around as in cyclic shift S. Every circulant is Col + CIS + ... + Cn_lSn - l . Cx = convolution c * x. Eigenvectors in F.
Diagonalizable matrix A.
Must have n independent eigenvectors (in the columns of S; automatic with n different eigenvalues). Then S-I AS = A = eigenvalue matrix.
Exponential eAt = I + At + (At)2 12! + ...
has derivative AeAt; eAt u(O) solves u' = Au.
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.
Hilbert matrix hilb(n).
Entries HU = 1/(i + j -1) = Jd X i- 1 xj-1dx. Positive definite but extremely small Amin and large condition number: H is ill-conditioned.
Identity matrix I (or In).
Diagonal entries = 1, off-diagonal entries = 0.
Left inverse A+.
If A has full column rank n, then A+ = (AT A)-I AT has A+ A = In.
Linear transformation T.
Each vector V in the input space transforms to T (v) in the output space, and linearity requires T(cv + dw) = c T(v) + d T(w). Examples: Matrix multiplication A v, differentiation and integration in function space.
The pivot row j is multiplied by eij and subtracted from row i to eliminate the i, j entry: eij = (entry to eliminate) / (jth pivot).
A directed graph that has constants Cl, ... , Cm associated with the edges.
In each column, choose the largest available pivot to control roundoff; all multipliers have leij I < 1. See condition number.
Permutation matrix P.
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.
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.
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.
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.
Simplex method for linear programming.
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!
Skew-symmetric matrix K.
The transpose is -K, since Kij = -Kji. Eigenvalues are pure imaginary, eigenvectors are orthogonal, eKt is an orthogonal matrix.
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.
Vandermonde matrix V.
V c = b gives coefficients of p(x) = Co + ... + Cn_IXn- 1 with P(Xi) = bi. Vij = (Xi)j-I and det V = product of (Xk - Xi) for k > i.