Problem 14 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The constants are \( c_1 = -1 \), \( c_2 = \frac{5}{2} \), and \( c_3 = -\frac{1}{2} \).

Step 1: Identify Initial Condition

We are given the initial condition \( \mathbf{X}(0) = \begin{pmatrix} 1 \ 3 \ 0 \end{pmatrix} \). This tells us the state of the system at \( t = 0 \). We'll use this information to find \( c_1 \), \( c_2 \), and \( c_3 \).

Step 2: Setup the Equation for \( \mathbf{X}(0) \)

Let's substitute \( t = 0 \) into the equation for \( \mathbf{X}(t) \):\[ \mathbf{X}(0) = c_1 \begin{pmatrix} 5 \ -3 \ 2 \end{pmatrix} e^{0} + c_2 \begin{pmatrix} 2 \ 0 \ 1 \end{pmatrix} e^{0} + c_3 \begin{pmatrix} -2 \ 0 \ 1 \end{pmatrix} e^{0} \] Given that \( e^{0} = 1 \), this simplifies to:\[ \mathbf{X}(0) = c_1 \begin{pmatrix} 5 \ -3 \ 2 \end{pmatrix} + c_2 \begin{pmatrix} 2 \ 0 \ 1 \end{pmatrix} + c_3 \begin{pmatrix} -2 \ 0 \ 1 \end{pmatrix} \].

Step 3: Formulate the System of Equations

We equate the expression for \( \mathbf{X}(0) \) with the initial condition:\[ c_1 \begin{pmatrix} 5 \ -3 \ 2 \end{pmatrix} + c_2 \begin{pmatrix} 2 \ 0 \ 1 \end{pmatrix} + c_3 \begin{pmatrix} -2 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 1 \ 3 \ 0 \end{pmatrix} \]This gives us three equations:1. \( 5c_1 + 2c_2 - 2c_3 = 1 \)2. \( -3c_1 = 3 \)3. \( 2c_1 + c_2 + c_3 = 0 \).

Step 4: Solve for \( c_1 \)

From the second equation, we have:\[ -3c_1 = 3 \]Solving for \( c_1 \), we divide both sides by -3:\[ c_1 = -1 \].

Step 5: Substitute \( c_1 \) and Solve for \( c_2 \) and \( c_3 \)

With \( c_1 = -1 \), substitute back into the first and third equations:1. \( 5(-1) + 2c_2 - 2c_3 = 1 \) simplifies to \( -5 + 2c_2 - 2c_3 = 1 \) leading to \( 2c_2 - 2c_3 = 6 \), or \( c_2 - c_3 = 3 \).2. \( 2(-1) + c_2 + c_3 = 0 \) simplifies to \( -2 + c_2 + c_3 = 0 \) leading to \( c_2 + c_3 = 2 \).We have the system:\[ c_2 - c_3 = 3 \]\[ c_2 + c_3 = 2 \].

Step 6: Solve the System for \( c_2 \) and \( c_3 \)

Add the two equations to eliminate \( c_3 \):\[ (c_2 - c_3) + (c_2 + c_3) = 3 + 2 \]\[ 2c_2 = 5 \]Thus, \( c_2 = \frac{5}{2} \).Subtract the second equation from the first to find \( c_3 \):\[ (c_2 - c_3) - (c_2 + c_3) = 3 - 2 \]\[ -2c_3 = 1 \]Thus, \( c_3 = -\frac{1}{2} \).

Step 7: Verify Solution

Insert \( c_1 = -1 \), \( c_2 = \frac{5}{2} \), \( c_3 = -\frac{1}{2} \) back into the original equations to ensure they satisfy all conditions:1. \( 5(-1) + 2\left(\frac{5}{2}\right) - 2\left(-\frac{1}{2}\right) = 1 \) holds true.2. \( -3(-1) = 3 \) holds true.3. \( 2(-1) + \frac{5}{2} - \frac{1}{2} = 0 \) holds true. The solutions are correct.

Key Concepts

Determinant of a Matrix

In linear algebra, the determinant of a matrix is a special number that can be calculated from its elements. It provides essential information about the matrix, such as whether the matrix is invertible or not. An invertible matrix has a non-zero determinant. This concept is pivotal for solving systems of linear equations and finding eigenvalues and eigenvectors.
To compute a determinant, one uses a specific formula or method depending on the size of the matrix:

• For a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is computed as \( ad - bc \).
• For larger matrices, the process involves more complex calculations, often utilizing cofactor expansion or row reduction.
• A zero determinant signifies that the matrix is singular, meaning it does not have an inverse.

In the context of finding eigenvalues, like in the original exercise, the determinant of \( \mathbf{A} - \lambda \mathbf{I} \) must equal zero. This expression is called the characteristic polynomial of the matrix \( \mathbf{A} \), where \( \lambda \) represents the eigenvalues.

Systems of Linear Equations

Systems of linear equations are collections of equations that involve the same set of variables. These systems can be solved using various methods, and they play a crucial role in many applications of linear algebra.
In simple terms, each equation in a system represents a line, plane, or hyperplane, depending on the number of variables. The solution to the system is the point(s) where all these geometrical objects intersect. Sometimes, there are no solutions, one solution, or infinitely many solutions.
To solve these systems, we can use methods such as:

• Substitution: Solve one equation for one variable and substitute it into the others.
• Elimination: Add or subtract equations to eliminate variables step by step.
• Matrix methods: Use matrices and operations like row reduction or finding the inverse to solve the system efficiently.

In the exercise, finding constants \( c_1, c_2, \text{and} \ c_3 \) involves forming a system of equations, which is solved to match a given initial condition. This showcases the practical use of linear algebra methods in finding specific solutions that satisfy the entire system.

Linear Algebra

Linear algebra is a vast field of mathematics that focuses on vectors, vector spaces, linear transformations, and matrices. It underpins many areas of mathematics and its applications are pervasive across science and engineering.
Here are some key topics within linear algebra:

• Vector spaces: Collections of vectors that can be added together and multiplied by scalars to produce another vector.
• Linear transformations: Functions that map vectors to vectors in a way that preserves vector addition and scalar multiplication.
• Matrices: Rectangular arrays of numbers used to represent linear transformations and solve systems of linear equations.
• Eigenvalues and eigenvectors: For a matrix \( \mathbf{A} \), an eigenvector is a non-zero vector that only changes by a scalar factor when the matrix is applied. The corresponding eigenvalue is the scalar.

Understanding these concepts provides the tools needed to solve complex problems such as those presented in the exercise. Linear algebra is fundamental for disciplines such as computer graphics, machine learning, and natural sciences, where systems of equations and transformations frequently occur.