Problem 17 - Differential Equations and Linear Algebra

Question

Determine the general solution to the linear system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\). \(\left[\begin{array}{rrr}-7 & -6 & -7 \\ -3 & -3 & -3 \\ 7 & 6 & 7\end{array}\right]\) [Hint: The eigenvalues of \(A \text { are } \lambda=0,-3 .]\)

Step-by-Step Solution

Answer

The general solution to the linear system \(\mathbf{x}'=A\mathbf{x}\) for the given matrix \(A\) is: \[ \mathbf{x}(t) = c_1\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} + c_2 e^{-3t}\begin{pmatrix} 3 \\ 3 \\ -2 \end{pmatrix}, \] where \(c_1\) and \(c_2\) are constants depending on the initial conditions.

Step 1: Eigenvalues of A

The hint given tells us that the eigenvalues of matrix \(A\) are \(\lambda=0\) and \(\lambda=-3\). We can move on to the next step since the eigenvalues are given.

Step 2: Find Eigenvectors for each Eigenvalue

To find the eigenvectors for each eigenvalue, we need to solve the equation \((A-\lambda I)\mathbf{x}=0\), where \(I\) is the identity matrix and \(\lambda\) is the eigenvalue. For \(\lambda=0\): Solving the equation \((A-0I)\mathbf{x}=0\) gives us the following system of linear equations: \[ \begin{cases} -7x_1 - 6x_2 - 7x_3 = 0 \\ -3x_1 - 3x_2 - 3x_3 = 0 \\ 7x_1 + 6x_2 + 7x_3 = 0 \end{cases} \] We can simplify it to: \[ \begin{cases} x_1 + x_2 + x_3 = 0 \\ x_2 + x_3 = 0 \end{cases} \] So, the eigenvector for \(\lambda=0\) can be written as: \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\) For \(\lambda=-3\): Solving the equation \((A-(-3)I)\mathbf{x}=0\) gives us the following system of linear equations: \[ \begin{cases} (-4)x_1 - 6x_2 - 7x_3 = 0 \\ -3x_1 - 6x_2 - 3x_3 = 0 \\ 7x_1 + 6x_2 + 10x_3 = 0 \end{cases} \] We can simplify it to: \[ \begin{cases} 2x_2 + 3x_3 = 0 \\ 7x_1 + 10x_3 = 0 \end{cases} \] So, the eigenvector for \(\lambda=-3\) can be written as: \(\mathbf{v}_2 = \begin{pmatrix} 3 \\ 3 \\ -2 \end{pmatrix}\)

Step 3: Write the general solution

Now that we have found the eigenvectors corresponding to the eigenvalues, we can write the general solution to the linear system using the exponential matrix function. The general solution can be written as: \[ \mathbf{x}(t) = c_1 e^{0t}\mathbf{v}_1 + c_2 e^{-3t}\mathbf{v}_2 = c_1\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} + c_2 e^{-3t}\begin{pmatrix} 3 \\ 3 \\ -2 \end{pmatrix}, \] where \(c_1\) and \(c_2\) are constants depending on the initial conditions.

Key Concepts

Eigenvalues

Eigenvalues are special numbers that reveal quite a lot about the characteristics of a matrix, particularly in the context of linear transformations. In our case, the original problem provided the eigenvalues, \[\lambda = 0 \text{ and } \lambda = -3.\] These values tell us about the nature of transformations that matrix \(A\) can apply to vectors.

• An eigenvalue of zero generally indicates that there's a stationary state or dimension where the transformation essentially "stops" or becomes ineffective.
• Negative eigenvalues, like our \(\lambda = -3\), suggest transformations that reverse direction and scale.

These eigenvalues are crucial because they influence stability and behavior over time in dynamic systems, like our linear system of differential equations.

Eigenvectors

For each eigenvalue, there's at least one corresponding eigenvector that lies along a direction in which the transformation represented by the matrix only stretches, compresses, or flips but does not alter direction.
The eigenvectors in this specific problem were determined by solving the matrix equation \((A - \lambda I)\mathbf{x} = 0\), where \(I\) is the identity matrix and \(\lambda\) denotes our eigenvalue.
For \(\lambda = 0\), the eigenvector \(\mathbf{v}_1\) was found to be:\[\mathbf{v}_1 = \begin{pmatrix} 1 \ -1 \ 1 \end{pmatrix}.\]
For \(\lambda = -3\), the eigenvector \(\mathbf{v}_2\) turned out to be:\[\mathbf{v}_2 = \begin{pmatrix} 3 \ 3 \ -2 \end{pmatrix}.\]
These vectors form the basis for constructing the general solution, indicating the paths taken by the system under the transformation.

General Solution

The general solution to a system of linear differential equations provides the complete description of all possible states and behaviors of the system over time. It is constructed using the eigenvalues and eigenvectors determined previously.
The general solution is expressed using the formula:\[\mathbf{x}(t) = c_1 e^{0t} \mathbf{v}_1 + c_2 e^{-3t} \mathbf{v}_2, \] where \(c_1\) and \(c_2\) are constants determined by initial conditions of the system.

• The exponential terms \(e^{\lambda t}\) account for the growth or decay influence of each eigenvalue.
• The accompanying eigenvectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\) indicate the direction of movement in the system.

This general form allows one to understand how solutions evolve over time, emphasizing how each part contributes to the overall behavior of the system.

Exponential Matrix Function

The exponential matrix function is a critical tool used to solve systems of linear differential equations, which are prevalent in engineering and physical sciences.
It is given by the term \(e^{At}\), bridging the linear algebra involved with linear differential equations.

• This function effectively transforms the matrix through an operation similar to scalar exponentiation, extending these concepts to matrices.
• The resulting matrix describes the system's behavior over time, capturing both rotation (complex parts) and scaling (real parts) of vectors.

In our problem, it allows the general solution to be elegantly stated in terms of exponential growth or decay, contingent upon the eigenvalues. It clarifies how the system will evolve from any initial state, providing a complete description encompassing all possible outcomes as time progresses.