Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# 0 0 a 1 0 26 a the inverse of e at is e at at to see

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: t as in 3 (a) If every column of A adds to zero, this means that the rows add to the zero row. So the rows are dependent, and A is singular, and  D 0 is an eigenvalue.   2 3 (b) The eigenvalues of A D are 1 D 0 with eigenvector x 1 D .3; 2/ and 2 3 2 D 5 (to give   D 5) with x D .1; 1/. Then the usual 3 steps: trace 2 4 3 1 1. Write u.0/ D as C D x1 C x2 1 2 1 2. Follow those eigenvectors by e 0t x 1 and e 5t x 2 3. The solution u.t / D x 1 C e 5t x 2 has steady state x 1 D .3; 2/.   1 1 4 d.v C w/=dt D .w v / C .v w / D 0, so the total v C w is constant. A D 1 1   1 D 0 1 1 v.1/ D 20 C 10e 2 v.1/ D 20 has with x 1 D , x2 D ; 2 D 2 1 1 w.1/ D 20 10e 2 w.1/ D 20 Solutions to Exercises 65    v 1 1 D has  D 0 and C2: v.t / D 20 C 10e 2t 1 as t ! 1. w 1 1   a1 6 AD has real eigenvalues a C 1 and a 1. These are both negative if a < 1, 1a   b 1 0 and the solutions of u D Au approach zero. B D has complex eigenvalues 1b b C i and b i . These have negative real parts if b < 0, and all solutions of v0 D B v approach zero. d 5 dt 7 A projection matrix has eigenvalues  D 1 and  D 0. Eigenvectors P x D x ﬁll the subspace that P projects onto: here x D .1; 1/. Eigenvectors P x D 0 ﬁll the perpendicular subspace: here x D .1; 1/. For the solution to u0 D P u,        3 2 1 1 1 t2 0t u.0/ D D C u.t / D e Ce approaches : 1 2 1 2 1 1 8 9 10 11 12     6 2 2 1 has 1 D 5, x 1 D , 2 D 2, x 2 D ; rabbits r.t / D 20e 5t C 10e 2t , 2 1 1 2 w.t / D 10e 5t C 20e 2t . The ratio of rabbits to wolves approaches 20=10; e 5t dominates.        4 1 1 1 4 cos t it 1 it (a) D2 C2 . (b) Then u.t / D 2e C2e D . 0 i i i i 4 sin t    0      dy y 01 y 01 D .AD has det.A I / D 2 5 4 D 0. 0D y 00 4 5 y0 45 dt y Directly substituting y D e t into y 00 D 5y 0 C 4y also gives 2 D 5 C 4 and the same p two values of . Those values are 1 .5 ˙ 41/ by the quadratic formula. 2          01 1t y.t / 1t y.0/ e At D I C t C zeros D . Then D 00 01 y 0 .t / 0 1 y 0 .0/   0 y.0/ C y .0/t . This y.t / D y.0/ C y 0 .0/t solves the equation. y 0 .0/   01 AD has trace 6, det 9,  D 3 and 3 with one independent eigenvector .1; 3/. 96 13 (a) y.t / D cos 3t and sin 3t  solve y 00D 9y . It is 3 cos 3t that starts with y.0/ D 3 01 and y 0 .0/ D 0. (b) A D has det D 9:  D 3i and 3i with x D .1; 3i / 90     1 1 3 cos 3t 3 3it 3 3it C 2e D . and .1; 3i /. Then u.t / D 2 e 3i 3i 9 sin 3t 14 When A is skew-symmetric, ku.t /k D ke At u.0/k is ku.0/k. So e At is orthogonal.     4 4 t1 t0 15 up D 4 and u.t / D ce C 4; up D and u.t / D c1 e C c2 e C . 2 t 1 2 t 16 Substituting u D e ct v gives ce ct v D Ae ct v .A cI / 1 e ct b or .A c I /v D b or v D b D particular solution. If c is an eigenvalue then A c I is not invertible. Solutions to Exercises 66       0 10 11 17 (a) (b) (c) . These show the unstable cases 1 01 11 (a) 1 < 0 and 2 > 0 (b) 1 > 0 and 2 > 0 (c)  D a ˙ i b with a > 0 1 0 1 1 1 1 18 d=dt .e At / D A C A2 t C 2 A3 t 2...
View Full Document

## This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

Ask a homework question - tutors are online