 10.17.1: Suppose that a certain animal population is divided into two age cl...
 10.17.2: Find the characteristic polynomial of a general Leslie matrix given...
 10.17.3: (a) Show that the positive eigenvalue of a Leslie matrix is always ...
 10.17.4: Show that the right side of Equation 10 is , where c is the first e...
 10.17.5: Show that the net reproduction rate R, defined by 17, can be interp...
 10.17.6: Show that a population is eventually decreasing if and only if its ...
 10.17.7: Calculate the net reproduction rate of the animal population in Exa...
 10.17.8: (For readers with a hand calculator) Calculate the net reproduction...
 10.17.9: (For readers who have read Section 10.1Section 10.3) Prove Theorem ...
 10.17.T1: Consider the sequence of Leslie matrices (a) Use a computer to show...
 10.17.T2: Consider the sequence of Leslie matrices where , and . (a) Choose a...
 10.17.T3: Suppose that a population of mice has a Leslie matrix L over a 1mo...
Solutions for Chapter 10.17: AgeSpecific Population Growth
Full solutions for Elementary Linear Algebra: Applications Version  10th Edition
ISBN: 9780470432051
Solutions for Chapter 10.17: AgeSpecific Population Growth
Get Full SolutionsThis textbook survival guide was created for the textbook: Elementary Linear Algebra: Applications Version, edition: 10. This expansive textbook survival guide covers the following chapters and their solutions. Elementary Linear Algebra: Applications Version was written by and is associated to the ISBN: 9780470432051. Since 12 problems in chapter 10.17: AgeSpecific Population Growth have been answered, more than 13846 students have viewed full stepbystep solutions from this chapter. Chapter 10.17: AgeSpecific Population Growth includes 12 full stepbystep solutions.

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

Cofactor Cij.
Remove row i and column j; multiply the determinant by (I)i + j •

Distributive Law
A(B + C) = AB + AC. Add then multiply, or mUltiply then add.

Exponential eAt = I + At + (At)2 12! + ...
has derivative AeAt; eAt u(O) solves u' = Au.

Graph G.
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.

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.

Lucas numbers
Ln = 2,J, 3, 4, ... satisfy Ln = L n l +Ln 2 = A1 +A~, with AI, A2 = (1 ± /5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = O.

Markov matrix M.
All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.

Multiplier eij.
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).

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.

Normal matrix.
If N NT = NT N, then N has orthonormal (complex) eigenvectors.

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

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.

Row picture of Ax = b.
Each equation gives a plane in Rn; the planes intersect at x.

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

Special solutions to As = O.
One free variable is Si = 1, other free variables = o.

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

Symmetric matrix A.
The transpose is AT = A, and aU = a ji. AI is also symmetric.

Vector addition.
v + w = (VI + WI, ... , Vn + Wn ) = diagonal of parallelogram.