Problem 15 - Advanced Engineering Mathematics

Question

Verify that the vector \(\mathbf{X}\) is a solution of the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rrr} 1 & 2 & 1 \\ 6 & -1 & 0 \\ -1 & -2 & -1 \end{array}\right) \mathbf{x} ; \quad \mathbf{X}=\left(\begin{array}{r} 1 \\ 6 \\ -13 \end{array}\right) $$

Step-by-Step Solution

Answer

Yes, \( \mathbf{X} \) is a solution as \( \mathbf{X}' = A\mathbf{X} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \).

Step 1: Understand the Problem Statement

We are given a system of first-order differential equations expressed in matrix form: \[ \mathbf{X}' = A\mathbf{x} \] where \[ A = \begin{pmatrix} 1 & 2 & 1 \ 6 & -1 & 0 \ -1 & -2 & -1 \end{pmatrix} \] and \( \mathbf{X} = \begin{pmatrix} 1 \ 6 \ -13 \end{pmatrix} \). Our task is to verify whether \( \mathbf{X} \) is a solution, which means checking if \( \mathbf{X}' = A \mathbf{X} \) holds true.

Step 2: Compute the Product A*X

To verify \( \mathbf{X}' = A \mathbf{x} \), calculate the matrix product \( A\mathbf{X} \): \[ A\mathbf{X} = \begin{pmatrix} 1 & 2 & 1 \ 6 & -1 & 0 \ -1 & -2 & -1 \end{pmatrix} \begin{pmatrix} 1 \ 6 \ -13 \end{pmatrix} \] This results in: \[ A\mathbf{X} = \begin{pmatrix} 1 \times 1 + 2 \times 6 + 1 \times (-13) \ 6 \times 1 + (-1) \times 6 + 0 \times (-13) \ -1 \times 1 + (-2) \times 6 + (-1) \times (-13) \end{pmatrix} \] \[ = \begin{pmatrix} 1 + 12 - 13 \ 6 - 6 + 0 \ -1 - 12 + 13 \end{pmatrix} \] \[ = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} \]

Step 3: Compare A*X with X'

Now, compare the result of \( A\mathbf{X} \) with \( \mathbf{X}' \). According to the problem statement, for \( \mathbf{X} \) to be a solution, the derivative \( \mathbf{X}' \) should equal the product \( A\mathbf{X} \). Since we found \( A\mathbf{X} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} \), this implies \( \mathbf{X}' = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} \). Therefore, \( \mathbf{X} \) is the trivial solution where its derivative is zero.

Key Concepts

Matrix Algebra

Matrix algebra plays a crucial role when dealing with systems of linear equations, especially in the context of differential equations. Here, matrices are employed to represent and solve these systems efficiently.
Matrix multiplication is particularly significant as it is used to determine solutions and verify them. This involves the multiplication of a matrix by a vector or another matrix, which results in a new matrix, allowing one to assess the influence of a transformation on the vector.

• In our case, we have matrix \( A \) and vector \( \mathbf{X} \). The product \( A\mathbf{X} \) determines the result of applying the transformation represented by \( A \) to the vector \( \mathbf{X} \).
• Matrix multiplication is not commutative, meaning \( A\mathbf{X} \) is generally not the same as \( \mathbf{X}A \).
• Each element of the resulting vector or matrix is calculated by taking the dot product of the corresponding row of the first matrix and the column of the second.

Understanding these concepts of matrix algebra is crucial in not only solving linear systems analytically but in leveraging computational software to manage larger, more complex systems.

Vector Calculus

Vector calculus expands upon traditional calculus by applying similar concepts to vector-valued functions. These functions have variables and produce vectors as outputs.
In the context of differential equations, especially those expressed in matrix form, vectors represent dynamic states or solutions over time.

• When given \( \mathbf{X} = \begin{pmatrix} 1 \ 6 \ -13 \end{pmatrix} \), this vector can be treated component-wise for calculations like derivatives and products.
• The derivation of a vector, like \( \mathbf{X}' \), entails finding the time rate of change of each component individually, leading to another vector of the same dimension.

Vector calculus is powerful when applied to fields like physics and engineering, where multiple directions of influence are assessed simultaneously, making it a pivotal tool in solving differential equations that describe such systems.

Linear Systems

Linear systems in mathematics encompass equations and their solutions where each equation or condition expresses a linear relationship between variables. These systems are particularly vital in fields that model real-world behaviors.
In our exercise, the system of differential equations can be expressed in a compact matrix form \( \mathbf{X}' = A\mathbf{x} \), where:

• \( A \) is a constant matrix representing coefficients of linear equations.
• \( \mathbf{x} \) is a vector of variables whose dynamics we need to understand.

Linear systems are advantageous for their predictability and solvability. Using matrix techniques, such as row operations and eigenvalue analysis, one can derive solutions and understand system behavior. These systems, when mapped into matrix form, simplify identifying dependencies or redundancies among equations, thereby streamlining solution methods.