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Math  /  Algebra

QuestionThere are ten problems. Solve all problems. You may use the eigenvalues and the linearly independent eigenvectors which you found in Homework 8 to solve the problems in this assignment. However, you must show all supporting work (except for the work to find the eigenvalues and the eigenvectors).
Let α≠0\alpha \neq 0 and β≠0\beta \neq 0 be real nonzero constants. Define the following real symmetric matrices (3) A3=(α−β−β−β−βα−β−β−β−βα−β−β−β−βα)A_{3}=\left(\begin{array}{cccc} \alpha & -\beta & -\beta & -\beta \\ -\beta & \alpha & -\beta & -\beta \\ -\beta & -\beta & \alpha & -\beta \\ -\beta & -\beta & -\beta & \alpha \end{array}\right)
The computations show that there are two cipenyalues λ1=α−βρ2λ2=α+β\lambda_{1}=\alpha-\beta \rho_{2} \quad \lambda_{2}=\alpha+\beta
Moreover, the bases of the eigenspaces are given by, respectively B1={(1111)} for the eigenvalue λ1=α−3β,B2={(11−1−1),(1−1−11),(1−11−1)}, for the eigenvalue λ2−a+B\begin{array}{l} \mathcal{B}_{1}=\left\{\left(\begin{array}{l} 1 \\ 1 \\ 1 \\ 1 \end{array}\right)\right\} \text { for the eigenvalue } \lambda_{1}=\alpha-3 \beta, \\ \mathcal{B}_{2}=\left\{\left(\begin{array}{c} 1 \\ 1 \\ -1 \\ -1 \end{array}\right),\left(\begin{array}{c} 1 \\ -1 \\ -1 \\ 1 \end{array}\right),\left(\begin{array}{c} 1 \\ -1 \\ 1 \\ -1 \end{array}\right)\right\} \text {, for the eigenvalue } \lambda_{2}-a+B \end{array}
Solve the following first order homogeneous systems of differential equations (3) ddtu=A3u\frac{\mathrm{d}}{\mathrm{d} t} \mathbf{u}=A_{3} \mathbf{u}

Studdy Solution

STEP 1

1. We are given a first-order homogeneous system of differential equations.
2. The matrix A3 A_3 is symmetric, and we have its eigenvalues and eigenvectors.
3. The solution can be expressed in terms of the eigenvalues and eigenvectors of A3 A_3 .

STEP 2

1. Verify the eigenvalues and eigenvectors.
2. Express the general solution using the eigenvectors.
3. Solve the differential equation using the general solution.

STEP 3

Verify the given eigenvalues and eigenvectors for the matrix A3 A_3 .
Given eigenvalues: - λ1=α−3β\lambda_1 = \alpha - 3\beta - λ2=α+β\lambda_2 = \alpha + \beta
Given bases for eigenspaces: - For λ1\lambda_1: B1={(1111)}\mathcal{B}_1 = \left\{\begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}\right\} - For λ2\lambda_2: B2={(11−1−1),(1−1−11),(1−11−1)}\mathcal{B}_2 = \left\{\begin{pmatrix} 1 \\ 1 \\ -1 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \\ -1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \\ 1 \\ -1 \end{pmatrix}\right\}

STEP 4

Express the general solution u(t) \mathbf{u}(t) using the eigenvectors and eigenvalues.
The general solution is a linear combination of solutions corresponding to each eigenvalue:
u(t)=c1eλ1t(1111)+c2eλ2t(11−1−1)+c3eλ2t(1−1−11)+c4eλ2t(1−11−1) \mathbf{u}(t) = c_1 e^{\lambda_1 t} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} + c_2 e^{\lambda_2 t} \begin{pmatrix} 1 \\ 1 \\ -1 \\ -1 \end{pmatrix} + c_3 e^{\lambda_2 t} \begin{pmatrix} 1 \\ -1 \\ -1 \\ 1 \end{pmatrix} + c_4 e^{\lambda_2 t} \begin{pmatrix} 1 \\ -1 \\ 1 \\ -1 \end{pmatrix}
where c1,c2,c3, c_1, c_2, c_3, and c4 c_4 are constants determined by initial conditions.

STEP 5

Solve the differential equation using the general solution.
The solution to the system ddtu=A3u\frac{\mathrm{d}}{\mathrm{d} t} \mathbf{u} = A_3 \mathbf{u} is given by:
u(t)=c1e(α−3β)t(1111)+c2e(α+β)t(11−1−1)+c3e(α+β)t(1−1−11)+c4e(α+β)t(1−11−1) \mathbf{u}(t) = c_1 e^{(\alpha - 3\beta) t} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} + c_2 e^{(\alpha + \beta) t} \begin{pmatrix} 1 \\ 1 \\ -1 \\ -1 \end{pmatrix} + c_3 e^{(\alpha + \beta) t} \begin{pmatrix} 1 \\ -1 \\ -1 \\ 1 \end{pmatrix} + c_4 e^{(\alpha + \beta) t} \begin{pmatrix} 1 \\ -1 \\ 1 \\ -1 \end{pmatrix}
The solution to the system of differential equations is expressed in terms of the eigenvalues and eigenvectors of the matrix A3 A_3 .

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