Problem 52 - Advanced Engineering Mathematics

Question

In Problems 51 and 52, write the system in the form \(\mathbf{A X}=\mathbf{B}\). Use \(\mathbf{X}=\mathbf{A}^{-1} \mathbf{B}\) to solve the system for each matrix \(\mathbf{B}\). $$ \begin{aligned} &x_{1}+2 x_{2}+5 x_{3}=b_{1} \\ &2 x_{1}+3 x_{2}+8 x_{3}=b_{2} \\ &-x_{1}+x_{2}+2 x_{3}=b_{3} \\ &\mathbf{B}=\left(\begin{array}{r} -1 \\ 4 \\ 6 \end{array}\right), \quad \mathbf{B}=\left(\begin{array}{l} 3 \\ 3 \\ 3 \end{array}\right), \quad \mathbf{B}=\left(\begin{array}{r} 0 \\ -5 \\ 4 \end{array}\right) \end{aligned} $$

Step-by-Step Solution

Answer

For each \(\mathbf{B}\): \([-71, -110, 23]\), \([-12, -18, 6]\), \([26, 38, -12]\).

Step 1: Write the System in Matrix Form

The given system of equations can be written in the matrix form \(\mathbf{A} \mathbf{X} = \mathbf{B}\), where \(\mathbf{A}\) is the coefficient matrix, \(\mathbf{X}\) is the vector of variables, and \(\mathbf{B}\) is the constant vector. For this system:\[ \mathbf{A} = \begin{bmatrix} 1 & 2 & 5 \ 2 & 3 & 8 \ -1 & 1 & 2 \end{bmatrix}, \quad \mathbf{X} = \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} , \quad \mathbf{B} = \begin{bmatrix} b_1 \ b_2 \ b_3 \end{bmatrix} \]

Step 2: Compute the Inverse of Matrix A

Calculate the inverse of \(\mathbf{A}\). The inverse of a 3x3 matrix \(\mathbf{A}\) can be found using a formula that involves determinants and minors. After computation, the inverse is:\[ \mathbf{A}^{-1} = \begin{bmatrix} 5 & -2 & -9 \ 4 & -1 & -13 \ -1 & 0 & 3 \end{bmatrix} \]

Step 3: Solve for X using each B

Now use \( \mathbf{X} = \mathbf{A}^{-1} \mathbf{B} \) for each given \(\mathbf{B}\):1. For \(\mathbf{B} = \begin{bmatrix} -1 \ 4 \ 6 \end{bmatrix}\): \[ \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} 5 & -2 & -9 \ 4 & -1 & -13 \ -1 & 0 & 3 \end{bmatrix} \begin{bmatrix} -1 \ 4 \ 6 \end{bmatrix} = \begin{bmatrix} -71 \ -110 \ 23 \end{bmatrix} \]2. For \(\mathbf{B} = \begin{bmatrix} 3 \ 3 \ 3 \end{bmatrix}\): \[ \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} 5 & -2 & -9 \ 4 & -1 & -13 \ -1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 3 \ 3 \ 3 \end{bmatrix} = \begin{bmatrix} -12 \ -18 \ 6 \end{bmatrix} \]3. For \(\mathbf{B} = \begin{bmatrix} 0 \ -5 \ 4 \end{bmatrix}\): \[ \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} 5 & -2 & -9 \ 4 & -1 & -13 \ -1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0 \ -5 \ 4 \end{bmatrix} = \begin{bmatrix} 26 \ 38 \ -12 \end{bmatrix} \]

Key Concepts

Matrix Equations

Matrix equations are a concise way to represent a system of linear equations. An entire system of equations can be rewritten using matrices for easier manipulation and computation. In our exercise, we represent the system in the form \( \mathbf{A}\mathbf{X} = \mathbf{B} \).

• \( \mathbf{A} \) is the coefficient matrix that contains the coefficients of the variables from the system of equations.
• \( \mathbf{X} \) is the column matrix containing the variables \( x_1, x_2, x_3 \).
• \( \mathbf{B} \) is the column matrix of constants \( b_1, b_2, b_3 \).

By transforming equations into this matrix form, complex systems can be more easily solved, especially when combined with computational tools. This method leverages the power of linear algebra to find solutions for multiple equations simultaneously. Through these matrix operations, the solutions become more systematic and less prone to arithmetic errors.

Inverse of a Matrix

The inverse of a matrix is central to solving matrix equations like \( \mathbf{A} \mathbf{X} = \mathbf{B} \). To isolate \( \mathbf{X} \), we need \( \mathbf{A}^{-1} \), which is the inverse of \( \mathbf{A} \). The product of a matrix and its inverse yields the identity matrix, effectively simplifying the equation.
Finding the inverse of a matrix involves a specific process:

• Ensure the matrix is square (same number of rows and columns).
• The matrix must have a non-zero determinant.
• Use algebraic minors, cofactors, and the determinant to calculate the inverse.

For our exercise, the matrix \( \mathbf{A} \) has been calculated to have an inverse \( \mathbf{A}^{-1} \), using these techniques. Utilizing the inverse simplifies our solution steps, enabling direct computation of \( \mathbf{X} \) by multiplying \( \mathbf{A}^{-1} \) with \( \mathbf{B} \). This solution method is efficient for systems where \( \mathbf{A} \) is known and its inverse is computable.

Systems of Linear Equations

Systems of linear equations consist of multiple linear equations with the same set of variables. The goal is to find the values of these variables that satisfy all the equations simultaneously.
In linear algebra, these systems can be expressed briefly using matrix notation. Matrix methods, such as row reduction or using the inverse of a matrix, provide powerful tools for solving systems, particularly when dealing with numerous equations.
Some key characteristics of systems of linear equations include:

• They can be consistent (a solution exists) or inconsistent (no solutions).
• They might have one solution, no solutions, or infinitely many solutions.
• They are highly relevant in modeling real-world scenarios, like physics problems and statistical models.

Instead of solving the equations one by one, utilizing the matrix equation form \( \mathbf{AX} = \mathbf{B} \) streamlines the solution process. It allows us to apply mathematical strategies and computational tools more effectively, yielding quicker insights into the system’s behavior and solutions.