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solutions11_2011

# solutions11_2011 - Math 152 Spring 2011 Solutions to...

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Unformatted text preview: Math 152, Spring 2011 Solutions to Assignment #11 1. Find all eigenvalues and eigenvectors of the matrix A= −4 −3 6 5 Soln: det(A − λI ) = det −4 − λ −3 = λ2 − λ − 2 = (λ − 2)(λ + 1) 6 5−λ =⇒ eigenvalues are 2, −1 For the eigenvalue λ = 2 A − λI = A − 2I = For the eigenvalue λ = −1, A − λI = A + I = −3 −3 6 6 =⇒ eigenvectors c 1 ,c = 0 −1 −6 −3 6 3 =⇒ eigenvectors c 1 ,c = 0 −2 2. Let A be the matrix of the problem above. Denote by v1 and v2 the eigenvectors you found for A. ˆ ˆ ˆ ˆ (a) Let e1 and e2 be the standard basis vectors for R2 . Express e1 and e2 as linear combinations ˆ ˆ e1 = s1 v1 + s2 v2 e2 = t1 v1 + t2 v2 of v1 and v2 . (b) Recall that if v is an eigenvector of A with eigenvalue λ and if n is any integer, ˆ ˆ then An v = λn v. Use this in conjunction with part (a) to ﬁnd A10 e1 and A10 e2 . (c) Find A10 . Soln: (a) Let ˆ e1 = 1 0 ˆ e2 = 0 1 v1 = 1 −2 v2 = 1 −1 ˆ ˆ By inspection e1 = 2v2 − v1 , e2 = v2 − v1 . The mechanical way to ﬁnd the linear combinations is to solve the systems of equations s1 v1 + s2 v2 = 1 1 −2 −1 1 s1 = 0 s2 1 t1 v1 + t2 v2 = 1 1 −2 −1 0 t1 = 1 t2 and ﬁnd s1 = −1, s2 = 2, t1 = −1, t2 = 1. (b) ˆ A10 e1 = A10 − v1 + 2v2 = −A10 v1 + 2A10 v2 = −210 v1 + 2(−1)10 v2 = −1024 1 1 = +2 −1 −2 −1022 2046 ˆ A10 e2 = A10 − v1 + v2 = −A10 v1 + A10 v2 = −210 v1 + (−1)10 v2 = −1024 (c) ˆ ˆ A10 = A10 e1 A10 e2 = −1022 −1023 2046 2047 1 1 = + −1 −2 −1023 2047 3. Find all eigenvalues and eigenvectors of the matrix B= 1 −1 11 Soln: det(B − λI ) = det 1 − λ −1 = λ2 − 2λ + 2 = (λ − 1 − i)(λ − 1 + i) 1 1−λ =⇒ λ = 1 ± i For the eigenvalue λ = 1 − i, B − λI = For the eigenvalue λ = 1 + i B − λI = −i −1 1 −i =⇒ eigenvectors c 1 ,c = 0 −i i −1 1i =⇒ eigenvectors c 1 ,c = 0 i 4. Find a matrix for which 2, −1]T is an eigenvector of eigenvalue 2 and 1, −1]T is an eigenvector of eigenvalue 6. Soln 1: Call the matrix ab A= cd Then ab cd 2 2 =2 −1 −1 2 ⇐⇒ 2a − b = 4 2c − d = −2 and ab cd 1 1 =6 −1 −1 ⇐⇒ a−b= 6 c − d = −6 The solution to 2a − b = 4, a − b = 6 is a = −2, b = −8. The solution to 2c − d = −2, c − d = −6 is c = 4, d = 10. So the matrix is −2 −8 4 10 Soln 2: We are told that A Hence A A 1 0 0 1 =A =A 2 1 − −1 −1 2 1 −2 −1 −1 =2 =2 2 1 −2 −6 = −1 −1 4 2 1 −8 − 12 = −1 −1 10 2 2 =2 −1 −1 A 1 1 =6 −1 −1 So the matrix is −2 −8 4 10 5. In 1949 there was a study of social mobility in England and Wales. All occupations were classiﬁed into three classes U: professional, managerial, executive M: inspectional, supervisory, other non–manual, skilled manual L: semi–skilled and unskilled manual The study collected data relating the occupational classes of sons to that of their fathers (only males were considered.) It found, for example, that 5.4% of sons of M–class fathers ended up in U–class occupations. The full transition matrix was U M L U 0.448 0.054 0.011 T = M 0.484 0.699 0.503 L 0.068 0.247 0.486 Find a vector x, having positive components that add up to 1, that obeys T x = x. Such a vector is called an equilibrium or steady state. 3 Soln: We need to ﬁnd a vector x1 , x2 , x3 that satisﬁes the equations x1 + x2 + x3 = 1, (T − I )x = 0. The augmented matrix for this system is 1 1 1 1 −0.552 0.054 0.011 0 0.484 −0.301 0.503 0 0.068 0.247 −0.514 0 Row reducing (1) 1 1 1 1 (2) + 0.552(1) 0 0.606 0.563 0.552 0 −0.785 0.019 −0.484 (3) − 0.484(1) 0 0.179 −0.582 −0.068 (4) − 0.068(1) (1) 1 1 1 1 0 0.606 0.563 0.552 (2) (3) + 0.785 (2) 0 0 0.748 0.231 0.606 0 0 0 0 (4) + (3) + (2) Backsolving x3 = x2 x1 0.231 = 0.309 0.748 1 0.552 − 0.563 × 0.309 = 0.624 = 0.606 = 1 − 0.309 − 0.624 = 0.067 4 ...
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