Problem 20 - Differential Equations and Linear Algebra

Question

Determine the general solution to the linear system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\). \(\left[\begin{array}{rrr}7 & -2 & 2 \\ 0 & 4 & -1 \\ -1 & 1 & 4\end{array}\right]\) [Hint: The only eigenvalue of \(A \text { is } \lambda=5 .]\)

Step-by-Step Solution

Answer

The general solution to the linear system \(\mathbf{x}'=A\mathbf{x}\) with the given matrix \(A\) is: \[\mathbf{x}(t) = c_1 e^{5t} \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}\] where \(c_1\) is a constant that depends on the initial conditions.

Step 1: Confirming the given eigenvalue

We are given that the only eigenvalue of matrix A is λ = 5. Let's verify this by calculating the determinant of (A - λI) and setting it to zero: \[ \text{det}(A - λI) = \text{det}\left(\begin{array}{ccc}7-\lambda & -2 & 2 \\\ 0 & 4-\lambda & -1 \\\ -1 & 1 & 4-\lambda\end{array}\right)\] To find the determinant of the 3x3 matrix, expand along the first column: \[ \text{det}(A - λI) = (7-\lambda) \cdot |4-\lambda, -1, 1, 4-\lambda|\] Now, find the determinant of the 2x2 matrix: \[ \text{det}(A - λI) = (7-\lambda)((4-\lambda)^{2} - (-1)(1))\] Setting the determinant to zero and solving for λ: \[(7-\lambda)((4-\lambda)^{2}+1) = 0\] From this equation, we indeed find that the eigenvalue λ is equal to 5.

Step 2: Find the eigenvector(s) associated with the eigenvalue λ = 5

To find the eigenvectors associated with λ = 5, we will solve the following equation for the vector x: \[(A - 5I)x = \mathbf{0}\] Substitute λ = 5: \[\left(\begin{array}{ccc}2 & -2 & 2 \\\ 0 & -1 & -1 \\\ -1 & 1 & -1\end{array}\right)x = \mathbf{0}\] Now we need to solve this system of homogeneous linear equations. We can notice that the third row is the negative of the first row, so we can remove the last row. We are left with following system: \[\begin{cases} 2x_1 - 2x_2 + 2x_3 = 0 \\ -x_2 - x_3 = 0 \end{cases}\] The second equation implies that: \[x_3 = -x_2\] Now, substitute this relationship into the first equation: \[(2x_1 - 2x_2 + 2(-x_2)) = 0\] \[2x_1 = 4x_2\] We can rewrite this equation as: \[x_1 = 2x_2\] This gives us the relationship between the components of the eigenvector x.

Step 3: Construct the general solution

Using the relationships found in Step 2, we can rewrite the eigenvector x as: \[x = \begin{bmatrix} 2x_2 \\ x_2 \\ -x_2 \end{bmatrix} = x_2 \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}\] Now that we have the eigenvector, we can construct the general solution for the linear system: \[\mathbf{x}(t) = c_1 e^{5t} \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}\] Here, c_1 is a constant that depends on the initial conditions of the system, and this is the general solution to the given linear system.

Key Concepts

Eigenvalue Calculation

When dealing with a system of linear equations, especially while working with matrices, eigenvalues play an essential role in understanding the behavior of the system. Eigenvalue calculation starts with a square matrix A and involves finding a scalar \(\lambda\) such that when we multiply the matrix by a nonzero vector (eigenvector), the product is the same as if we had simply multiplied that vector by the scalar. The equation encapsulating this relationship is \(A\textbf{x} = \lambda\textbf{x}\).

To find the eigenvalues, we set up the equation \(\text{det}(A - \lambda I) = 0\), where \(I\) is the identity matrix of the same dimension as A. The determinant of the matrix \(A - \lambda I\) gives us a polynomial, and the roots of this polynomial are the eigenvalues. Calculating the eigenvalues is crucial because it informs us about the invariant directions under the transformation implied by matrix A.

In the provided exercise, we confirm that the eigenvalue \(\lambda = 5\) is indeed a valid solution by solving for \(\lambda\) in the characteristic polynomial that results from \(\text{det}(A - \lambda I) = 0\).

Eigenvector Determination

After we have computed the eigenvalues of the matrix, the next step is to find the corresponding eigenvectors. An eigenvector associated with a particular eigenvalue is a nonzero vector \(\textbf{x}\) that satisfies the equation \(A\textbf{x} = \lambda\textbf{x}\). To determine these eigenvectors, we solve the homogeneous equation \(\left(A - \lambda I\right)\textbf{x} = \textbf{0}\), where \(\lambda\) is an eigenvalue of A and \(\textbf{0}\) represents the zero vector.

The system of equations that results from this operation may have infinitely many solutions, and we often express these solutions in terms of free variables, indicating that any scalar multiples of these solutions are also eigenvectors. In the specific exercise, we take \(\lambda = 5\) and solve for \(\textbf{x}\), leading us to the eigenvectors that span the eigenspace corresponding to the eigenvalue \(\lambda\).

Homogeneous Linear Equations

Homogeneous linear equations are systems of linear equations where all of the constant terms are zero, which can be denoted as \(A\textbf{x} = \textbf{0}\). These systems have a guaranteed solution, which is the trivial solution \(\textbf{x} = \textbf{0}\). However, when the determinant of A is zero, there are also non-trivial solutions, meaning there are eigenvectors associated with the eigenvalue \(\lambda = 0\) in the case of eigenvalue problems.

The steps to find the general solution of a homogeneous system involve reducing the system to its simplest form through row operations and identifying the relationships among the variables. As seen in the exercise, after identifying that one equation was a scalar multiple of another, we were able to reduce the system allowing us to find the relationships between the variables \(x_1\), \(x_2\), and \(x_3\). Understanding homogeneous linear equations is fundamental in linear algebra since they appear frequently in different contexts, including the study of eigenvectors.