Question

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{lll}{3} & {2} & {4} \\ {2} & {0} & {2} \\ {4} & {2} & {3}\end{array}\right) \mathbf{x} $$

Short answer

Question: Write the general solution for the given system of equations: \(x' = \begin{pmatrix} 3 & 2 & 4 \\ 2 & 0 & 2 \\ 4 & 2 & 3 \end{pmatrix}x\). Answer: The general solution of the given system of equations is: \( \mathbf{x}(t) = c_1 \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}+c_2 e^{2t} \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}+c_3 e^{4t} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \).

Step-by-step solution

Compute the Eigenvalues of A

To find the eigenvalues of the matrix A, we first calculate the characteristic equation, given by \(|A-\lambda I|=0\), where \(\lambda\) represents the eigenvalues and \(I\) is the identity matrix. So, we have: $$ \text{det}\left(\begin{array}{ccc} 3-\lambda & 2 & 4 \\ 2 & -\lambda & 2 \\ 4 & 2 & 3-\lambda \end{array}\right)=0 $$ Expanding the determinant, we get the characteristic equation: \((3-\lambda)((-\lambda)(3-\lambda)-4)-2(2(3-\lambda)-8)+4(4-2^2)=0 \Rightarrow \lambda^3-6\lambda^2+8\lambda=0\) Next, let's find the roots of this equation, which will be the eigenvalues.

Find Eigenvalues of the Characteristic Equation

We have the characteristic equation: \( \lambda^3-6\lambda^2+8\lambda=0 \). Factor the equation: \(\lambda(\lambda^2-6\lambda+8) = 0\) Now solve for the roots: \(\lambda_1=0\), \(\lambda_2=2\), \(\lambda_3=4\) These are the eigenvalues of the matrix A.

Find Eigenvectors for Each Eigenvalue

Now, we will find the eigenvectors corresponding to each eigenvalue by solving the equation \((A-\lambda_i I) \mathbf{v}_i = 0\), where \(\lambda_i\) and \(\mathbf{v}_i\) are the eigenvalue and its corresponding eigenvector, respectively. Eigenvectors for \(\lambda_1=0\): \((A-0I)\mathbf{v}_1 = \left(\begin{array}{ccc}{3} & {2} & {4} \\ {2} & {0} & {2} \\ {4} & {2} & {3}\end{array}\right)\mathbf{v}_1 = 0\) Row reducing to echelon form, we get: \(\mathbf{v}_1=k_1\begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}\) Eigenvectors for \(\lambda_2=2\): \((A-2I)\mathbf{v}_2 = \left(\begin{array}{ccc}{1} & {2} & {4} \\ {2} & {-2} & {2} \\ {4} & {2} & {1}\end{array}\right)\mathbf{v}_2 = 0\) Row reducing to echelon form, we get: \(\mathbf{v}_2=k_2\begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}\) Eigenvectors for \(\lambda_3=4\): \((A-4I)\mathbf{v}_3 = \left(\begin{array}{ccc}{-1} & {2} & {4} \\ {2} & {-4} & {2} \\ {4} & {2} & {-1}\end{array}\right)\mathbf{v}_3 = 0\) Row reducing to echelon form, we get: \(\mathbf{v}_3=k_3\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\)

Form the General Solution

Now that we have the eigenvalues and the corresponding eigenvectors, we can write the general solution of the given system as: \(\mathbf{x}(t)=c_1 e^{0t} \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}+c_2 e^{2t} \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}+c_3 e^{4t} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\) Or $$ \mathbf{x}(t) = c_1 \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}+c_2 e^{2t} \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}+c_3 e^{4t} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} $$ This is the general solution of the given system of equations.

Key concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial when it comes to solving systems of differential equations.

Eigenvalues, often denoted by \(\lambda\), are scalars that, when multiplied by a specific non-zero vector called an eigenvector, result in a vector that's parallel to the original one. This concept is beautifully captured in the equation \( A\mathbf{v} = \lambda\mathbf{v} \), where \( A \) is a square matrix, and \( \mathbf{v} \) is its eigenvector.

To find the eigenvalues of a given matrix \( A \), we use the characteristic equation. Once the eigenvalues are obtained, we can find the corresponding eigenvectors by solving \( (A - \lambda I)\mathbf{v} = 0 \), where \( I \) is the identity matrix. An eigenvalue can have more than one linearly independent eigenvector associated with it, each forming an eigenspace.

In the context of our exercise, each eigenvalue \( \lambda_i \) has a corresponding eigenvector \( \mathbf{v}_i \) that satisfies the equation. These eigenvectors are essential pieces in constructing the general solution of the system of equations.

Characteristic Equation

The characteristic equation is the foundation in the process of finding eigenvalues for a matrix. It is derived from the determinant of the matrix modified by an identity matrix scaled by \(\lambda\), the eigenvalue.

Mathematically, if \( A \) is our matrix, the characteristic equation takes the form of \( |A - \lambda I| = 0 \), where \( |...| \) denotes the determinant. Expanding this determinant will give you a polynomial in terms of \( \lambda \), which is called the characteristic polynomial. The roots of this polynomial are the eigenvalues we are looking for.

Solving the characteristic equation can sometimes be simplified by factoring or using methods such as the Rational Root Theorem or synthetic division. In the given exercise, expanding the determinant gives us \( \lambda^3 - 6\lambda^2 + 8\lambda = 0 \), from which eigenvalues are found by factoring and solving for \( \lambda \).

Determinant of a Matrix

The determinant is a special scalar value that can be calculated from the elements of a square matrix. It provides important information about the matrix, such as whether the matrix is invertible (a non-zero determinant) or singular (a determinant of zero).

The calculation of the determinant varies with the size of the matrix: for a 2x2 matrix, it's a straightforward formula, while for larger matrices, it involves recursion and can become quite complex. Expansion by minors, cofactor expansion, and the use of the Leibniz formula are common methods for calculating determinants.

For the exercise in question, the determinant is used within the characteristic equation to solve for the eigenvalues of the matrix. In effect, the determinant's role here is pivotal in defining the nature of the solutions we obtain for the system of equations.

Matrix Exponential

The matrix exponential, denoted as \( e^{At} \), where \( A \) is a matrix and \( t \) is a scalar, is a fundamental concept in solving systems of linear differential equations. This function is akin to the scalar exponential function and it can be defined via its Taylor series expansion.

One of the most relevant applications of the matrix exponential is in finding the solution to linear systems of differential equations. When combined with the eigenvalues and eigenvectors of a matrix, it provides a way to construct the general solution of the system.

In our exercise, the term \( e^{\lambda t} \) in the general solution represents the matrix exponential. Each term of the solution is formed by an eigenvector scaled by the exponential of its corresponding eigenvalue multiplied by \( t \). This serves to illustrate how the system evolves over time, making the matrix exponential a powerful tool in understanding dynamic systems.