Problem 30 - Advanced Engineering Mathematics

Question

We have \(\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=-(\lambda+1)(\lambda-1)^{2}=0 .\) For \(\lambda_{1}=-1\) we obtain $$\mathbf{K}_{1}=\left(\begin{array}{r} -1 \\ 0 \\ 1 \end{array}\right)$$ For \(\lambda_{2}=1\) we obtain $$\mathbf{K}_{2}=\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right) \quad \text { and } \quad \mathbf{K}_{3}=\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right)$$ so that $$\mathbf{X}=c_{1}\left(\begin{array}{r} -1 \\ 0 \\ 1 \end{array}\right) e^{-t}+c_{2}\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right) e^{t}+c_{3}\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) e^{t}$$ If $$\mathbf{X}(0)=\left(\begin{array}{l} 1 \\ 2 \\ 5 \end{array}\right)$$ then \(c_{1}=2, c_{2}=3,\) and \(c_{3}=2\)

Step-by-Step Solution

Answer

The constants are already calculated: \( c_1 = 2, c_2 = 3, c_3 = 2 \).

Step 1: Understand the System of Equations

We are given the vector \( \mathbf{X} \) as a linear combination of specific vectors with exponential functions as multipliers. We need to solve for \( c_1, c_2, \) and \( c_3 \) given that \( \mathbf{X}(0) \) is a specific vector.

Step 2: Substitute Initial Condition

Substitute \( t = 0 \) into the expression for \( \mathbf{X} \) to find the equation: \( \mathbf{X}(0) = c_1 \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix} e^{0} + c_2 \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} e^{0} + c_3 \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} e^{0} \). This simplifies to \( \mathbf{X}(0) = \begin{pmatrix} c_2 - c_1 \ c_3 \ c_1 + c_2 \end{pmatrix} \).

Step 3: Set Up the Equation System

Convert \( \mathbf{X}(0) \) to match its given value \( \begin{pmatrix} 1 \ 2 \ 5 \end{pmatrix} \). This gives the equations: 1. \( c_2 - c_1 = 1 \) 2. \( c_3 = 2 \) 3. \( c_1 + c_2 = 5 \)

Step 4: Solve the Equations

- Use equation 2: \( c_3 = 2 \).- Substitute into equation 1: Matching to \( c_2 - c_1 = 1 \), solve for one variable in terms of the other.- Solve equation 3: From \( c_1 + c_2 = 5 \), use the expression from equation 1 to find the exact values of \( c_1 \) and \( c_2 \).

Step 5: Find Specific Values

Substituting \( c_1 = 2 \) and \( c_2 = 3 \) back into the equations confirms they satisfy all equations: - \( c_2 - c_1 = 3 - 2 = 1 \)- \( c_1 + c_2 = 2 + 3 = 5 \)- \( c_3 = 2 \) remains consistent. So \( c_1 = 2, c_2 = 3, \) and \( c_3 = 2 \).

Step 6: Final Check

Verify that the coefficients \( c_1, c_2, \) and \( c_3 \) satisfy the initial conditions. Substitute them back into the original matrix equation to ensure they provide \( \mathbf{X}(0) = \begin{pmatrix} 1 \ 2 \ 5 \end{pmatrix} \). Everything checks out.

Key Concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial for solving many linear algebra problems, especially those involving matrices. Given a square matrix \(\mathbf{A}\), an eigenvalue \(\lambda\) is a scalar that, when we subtract it times the identity matrix \(\mathbf{I}\), causes the resulting matrix \(\mathbf{A} - \lambda \mathbf{I}\) to be singular (its determinant is zero). This singularity indicates that the matrix cannot be inverted, and it's essentially where the concept of eigenvectors comes in.
If \(\mathbf{v}\) is an eigenvector corresponding to an eigenvalue \(\lambda\), then for the matrix \(\mathbf{A}\), we have the relationship:

• \(\mathbf{A}\mathbf{v} = \lambda\mathbf{v}\)

This equation means that applying the matrix to the vector \(\mathbf{v}\) scales the vector by the eigenvalue \(\lambda\). In our example, solving the equation \(\text{det}(\mathbf{A}-\lambda \mathbf{I}) = 0\) gives us the eigenvalues \(\lambda_1 = -1\) and \(\lambda_2 = 1\). The associated eigenvectors, such as \(\begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix}\), align with these eigenvalues.
Understanding these concepts helps us find solutions to matrix equations, which is foundational in various fields like physics and chemistry.

Matrix Exponentials

Matrix exponentials extend the concept of an exponential function to matrices. If \(\mathbf{A}\) is a matrix, the exponential \(e^{\mathbf{A}t}\) is defined in a way analogous to the function \(e^{at}\) for scalar \(a\). This is useful in solving systems of linear differential equations and appears in the study of dynamic systems.

• For a diagonal matrix \(\mathbf{D}\), the matrix exponential is computed as:
  
  \( e^{\mathbf{D}t} = \begin{pmatrix} e^{\lambda_1 t} & \cdots & 0 \ \vdots & \ddots & \vdots \ 0 & \cdots & e^{\lambda_n t} \end{pmatrix} \)

In our exercise, the matrix exponential transforms eigenvectors \(\begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}\), \(\begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix}\), and \(\begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}\) by scaling each component with an exponential factor based on the corresponding eigenvalue. The multiplication of the matrix exponential's result with each eigenvector reflects this growth or decay over time.
Thus, knowing how to compute and interpret matrix exponentials aids in modeling dynamic behaviors over time, making it a powerful tool in engineering and scientific computations.

System of Linear Equations

Systems of linear equations are at the heart of linear algebra, providing ways to solve for unknown variables based on relationships expressed via equations. A typical linear equation takes the form \(ax + by + cz = d\). A set of several such equations involving the same variables forms a system.
The essence of solving a system, whether by substitution, elimination, or matrix operations like Gaussian elimination, is to determine the values of variables that satisfy all equations simultaneously.
In the original exercise, we use initial conditions, specifically \(\mathbf{X}(0) = \begin{pmatrix} 1 \ 2 \ 5 \end{pmatrix}\), to solve for constants \(c_1, c_2\), and \(c_3\). This boils down to solving a system of linear equations derived from expressions involving eigenvectors weighted by these constants.

• The equations derived were:
  1. \(c_2 - c_1 = 1\)
  2. \(c_3 = 2\)
  3. \(c_1 + c_2 = 5\)

By applying substitution and elimination techniques, we find the precise values for the constants as \(c_1 = 2\), \(c_2 = 3\), and \(c_3 = 2\). This successful approach not only solves the immediate problem but also reinforces the broader applicability of linear systems in mathematical modeling and problem-solving.