Question

In Exercises \(1-16\) find the general solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} -3 & 3 & 1 \\ 1 & -5 & -3 \\ -3 & 7 & 3 \end{array}\right] \mathbf{y} $$

Short answer

Answer: The eigenvalues of matrix A are: λ₁ = -3, λ₂ = -3 + i, and λ₃ = -3 - i.

Step-by-step solution

Write down the given matrix A

Write down the matrix A from the given equation: $$ A = \left[\begin{array}{rrr} -3 & 3 & 1\\ 1 & -5 & -3\\ -3 & 7 & 3 \end{array}\right] $$

Find the eigenvalues of A

Find the eigenvalues by solving the characteristic equation \(|A - \lambda I| = 0\) where \(I\) is the identity matrix and \(\lambda\) represents the eigenvalues. The characteristic equation is given by: $$ \left|\begin{array}{ccc} -3-\lambda & 3 & 1\\ 1 & -5-\lambda & -3\\ -3 & 7 & 3-\lambda \end{array}\right| = 0 $$ After calculating the determinant, we get the characteristic equation as follows: $$ \lambda^3 + 5 \lambda^2 + 3 \lambda + 59 = 0 $$ The eigenvalues are \(\lambda_1 = -3\), \(\lambda_2 = -3 + i\), and \(\lambda_3 = -3 - i\).

Find the eigenvectors of A

For each eigenvalue, find the eigenvectors by solving the equation \((A - \lambda I) \mathbf{v} = 0\). For \(\lambda_1 = -3\): $$ \left[\begin{array}{ccc} 0 & 3 & 1\\ 1 & -2 & -3\\ -3 & 7 & 6 \end{array}\right] \mathbf{v_1} = 0 $$ After row reducing to echelon form, we find the eigenvector \(\mathbf{v_1} = \left[\begin{array}{r}1\\1\\-1\end{array}\right]\). For \(\lambda_2 = -3 + i\): $$ \left[\begin{array}{ccc} -i & 3 & 1\\ 1 & -2 - i & -3\\ -3 & 7 & 6 - i \end{array}\right] \mathbf{v}_2 = 0 $$ After row reducing to echelon form, we find the eigenvector \(\mathbf{v_2} = \left[\begin{array}{r}3 + 2i\\1 + 4i\\0\end{array}\right]\). For \(\lambda_3 = -3 - i\): $$ \left[\begin{array}{ccc} i & 3 & 1\\ 1 & -2 + i & -3\\ -3 & 7 & 6 + i \end{array}\right] \mathbf{v}_3 = 0 $$ After row reducing to echelon form, we find the eigenvector \(\mathbf{v_3} = \left[\begin{array}{r}3 - 2i\\1 - 4i\\0\end{array}\right]\).

Write down the general solution

Using the eigenvalues and eigenvectors found in previous steps, we can form the general solution as follows: $$ \mathbf{y}(t) = c_1 e^{-3t} \left[\begin{array}{r}1\\1\\-1\end{array}\right] + c_2 e^{(-3 + i)t}\left[\begin{array}{r}3 + 2i\\1 + 4i\\0\end{array}\right] + c_3 e^{(-3 - i)t}\left[\begin{array}{r}3 - 2i\\1 - 4i\\0\end{array}\right] $$ This is the general solution of the given system of linear ordinary differential equations.

Key concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial for solving linear systems of differential equations. Essentially, eigenvalues are scalars that reveal how matrices act on vectors called eigenvectors. To determine them, you compute the characteristic equation, but let's discuss that later. When you have a matrix \( A \) and an eigenvalue \( \lambda \), the relationship \( (A - \lambda I)\mathbf{v} = 0 \) should satisfy. Here, \( \mathbf{v} \) is the eigenvector associated with the eigenvalue \( \lambda \).

For instance, in the provided step-by-step solution, we found the eigenvalues \( \lambda_1 = -3 \), \( \lambda_2 = -3 + i \), and \( \lambda_3 = -3 - i \). Each eigenvalue corresponds to a specific direction in which the transformation confidently stretches or compresses space. Finding the eigenvectors involved solving the system for each eigenvalue:

• For \( \lambda_1 = -3 \), the eigenvector is \( \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix} \).
• For \( \lambda_2 = -3 + i \), the eigenvector is \( \begin{bmatrix} 3 + 2i \ 1 + 4i \ 0 \end{bmatrix} \).
• For \( \lambda_3 = -3 - i \), the eigenvector is \( \begin{bmatrix} 3 - 2i \ 1 - 4i \ 0 \end{bmatrix} \).

These eigenvectors provide directions in which the transformation takes action. By understanding these concepts, you can decipher how matrices interact with vectors, a fundamental aspect of solving differential equations.

Characteristic Equation

The characteristic equation is essential to understanding both the behavior of linear transformations and solving differential systems. To find eigenvalues, we solve the characteristic equation \( |A - \lambda I| = 0 \). Here, \( A \) is our matrix, \( \lambda \) are the eigenvalues, and \( I \) is the identity matrix, which matches the size of \( A \).

In solving the provided exercise, you first subtract \( \lambda \) from the diagonal of the matrix \( A \) and then compute the determinant of this new matrix. For the matrix in the problem, the steps were as follows:

• Substituted \( \lambda \) in the diagonal of matrix \( A \), forming a new matrix.
• Find the determinant of this changed matrix to get \( \lambda^3 + 5 \lambda^2 + 3 \lambda + 59 = 0 \).

This new equation represents the characteristic equation whose roots give the eigenvalues. Consequently, solving \( \lambda^3 + 5 \lambda^2 + 3 \lambda + 59 = 0 \) introduced the eigenvalues: \(-3 \), \(-3 + i\), and \(-3 - i \). Understanding how to manipulate and solve the characteristic equation is a pivotal step in linear algebra and differential equations.

General Solution of Differential Equations

Finding the general solution for a system of linear differential equations relies on understanding the relationship between eigenvalues and eigenvectors. Once you find these, the general solution encapsulates all potential behaviors of the system over time. The solution is expressed as a combination of exponential terms weighted by corresponding eigenvectors.

Referencing our original solution, here's how to employ the eigenvalues and eigenvectors:\[ \mathbf{y}(t) = c_1 e^{-3t} \begin{bmatrix}1\1\-1\end{bmatrix} + c_2 e^{(-3+i)t}\begin{bmatrix}3 + 2i\1 + 4i\0\end{bmatrix} + c_3 e^{(-3-i)t}\begin{bmatrix}3 - 2i\1 - 4i\0\end{bmatrix} \]

• \( c_1, c_2, c_3 \) are constants determined by initial conditions.
• The terms \( e^{\lambda t} \) show growth or decay (real part) and oscillation (imaginary part) over time.
• These exponential terms, when multiplied by their eigenvectors, depict all possible states of the system.

This comprehensive formula reveals how the system evolves with time, determining whether solutions grow, shrink or oscillate depending on the context. Understanding this concept provides the foundation to tackling more complex systems and examining their long-term behaviors.