Homework 7 Solution - Introduction to Dierential Equations...

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Unformatted text preview: Introduction to Diﬀerential Equations – Math 286 X1 Fall 2009 Homework 7 Solutions 1. Let 1 3 A= 2 4 , 1 1 1 −1 B= . Compute A + B. Compute AB and BA. Does AB = BA? Solution: We have A+B = 2 4 3 3 , 3 7 AB = −1 −1 , 2 0 0 2 4 6 −2 −2 BA = . We see that AB = BA. 2. Let A= 1 3 2 4 , B= 2 6 4 8 , BA = . Compute AB and BA. Does AB = BA? Solution: We compute AB = 2 6 4 8 , so AB = BA. 3. Let 1 2 2 1 A= . Find a matrix B so that AB = BA. Now ﬁnd a second one. Check that they work. Solution: There are several ways to go here. We could choose B = A, and of course if A and B are equal it does not matter in which order we multiply them. We could also choose B = I, since AI = IA = A. Pushing these ideas further, if we choose B = αA for any number α, then AB = A(αA) = αAA = αA2 , BA = (αA)A = αA2 . Similarly, we can choose B = αI, since AB = A(αI) = αAI = αA, BA = αIA = αA. 4. For each of the following, determine whether or not the matrix has an inverse, and if it does, compute it: 1 2 1 1 1 1 1 3 1 3 , , , , . 3 4 1 −1 2 2 3 1 −1 −3 Solution: 1 (a) Determinant is −2, so inverse exists. We have A−1 = 1 −2 4 −2 −3 1 = −2 1 3/2 −1/2 . = 1/2 1/2 1/2 −1/2 . (b) Determinant is −2, so inverse exists. We have A−1 = 1 −2 −1 −1 −1 1 (c) Determinant is 0, so no inverse. (d) Determinant is −8, so inverse exists. We have A−1 = 1 −8 1 −3 −3 1 −1/8 3/8 3/8 −1/8 = (e) Determinant is 0, so no inverse. 5. Solve the system x′ = −x1 + 3x2 , 1 x′ = −2x1 + 4x2 , 2 x1 (0) = 2, x2 (0) = 3. Solution: We write this in matrix form as x′ = Ax, A= −1 3 −2 4 . We need to ﬁnd the eigenvalues and eigenvectors of A. We have −1 − λ −2 A − λI = 3 4−λ and det(A − λI) = (−1 − λ)(4 − λ) + 6 = λ2 − 3λ + 2. The roots are λ = 1, 2. For λ1 = 1, we solve (P − I)x = −2 3 −2 3 x y = 0 0 , = 0 0 , which gives and we choose v1 = 3 2 −2x + 3y = 0, . For λ2 = 2, we solve (P − I)x = −3 3 −2 2 x y which gives −3x + 3y = 0, 2 . and we choose v2 = 1 1 . Thus our general solution is 3 2 x(t) = C1 et 1 1 + C2 e2t . Plugging in t = 0 gives x(0) = C1 3 2 + C2 1 1 2C1 + C2 3C1 + C2 = , so we solve 3C1 + C2 = 2, 2C1 + C2 = 3. This gives C1 = −1, C2 = 5, so our solution is −3et + 5e2t −2et + 5e2t 3 2 + 5e2t 1 1 = x′ = 1 3 1 −1 x, x(0) = −et . 6. Solve the system 1 −1 . Solution: We ﬁrst ﬁnd the eigenvalues and eigenvectors. The trace of the matrix is 0 and the determinant is −4, so the characteristic polynomial is λ2 + 4 = 0, which has roots λ = ±2. For λ1 = 2, we have A − 2I = −1 3 1 −3 x y 0 0 = , which gives −x + 3y = 0, and we choose v1 = 3 1 . For λ1 = −2, we have A + 2I = 3 3 1 1 x y = 0 0 , 1 −1 . which gives 3x + 3y = 0, and we choose v2 = 1 −1 . Thus our general solution is x(t) = C1 e2t 3 1 3 + C2 e−2t Plugging in t = 0 gives 3C1 + C2 C1 − C2 x(0) = so we solve , 3C1 + C2 = 1, C1 − C2 = −1. This gives C1 = 0, C2 = 1, so we have x(t) = e−2t 1 −1 . 7. Solve the system 1 2 x′ = 1 1 x, 1 0 x(0) = . Solution: We ﬁrst ﬁnd the eigenvalues and eigenvectors. The trace of the matrix is 2 and the determinant is −1, so the characteristic polynomial is λ2 − 2λ − 1 = 0, which has roots λ= For λ1 = 1 + √ 2, we have A − (1 + and we choose v1 = 1 √ 2 √ − 2 2 √ 2)I = which gives For λ1 = 1 − √ √ 4+4 = 1 ± 2. 2 2± 1 √ − 2 x y 0 0 = , √ − 2x + y = 0, . √ 2, we have A − (1 − which gives and we choose v2 = √ 2 2 √ 2)I = √ 1 √ − 2 1 √ 2 x y = 0 0 , 2x + y = 0, . Thus our general solution is x(t) = C1 e(1+ √ 2)t 1 √ 2 + C2 e(1− √ 2)t Plugging in t = 0 gives x(0) = √ C1 + C2 2(C1 − C2 ) 4 , 1 √ − 2 . so we solve C1 + C2 = 1, √ 2(C1 − C2 ) = 0. This gives C1 = 1/2, C2 = 1/2, so we have x(t) = √ √ 1 (1+ 2)t 2)t + e(1− √ ) 2 (e √ 2 (1+ 2)t − e(1− 2)t ) 2 (e 5 . ...
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