Problem 51 - Advanced Engineering Mathematics

Question

Write the system in the form \(\mathbf{A} \mathbf{X}=\mathbf{B}\). Use \(\mathbf{X}=\mathbf{A}^{-1} \mathbf{B}\) to solve the system for each matrix \(\mathbf{B}\). $$ \begin{aligned} &7 x_{1}-2 x_{2}=b_{1} \\ &3 x_{1}-2 x_{2}=b_{2} \\ &\mathbf{B}=\left(\begin{array}{l} 5 \\ 4 \end{array}\right), \quad \mathbf{B}=\left(\begin{array}{c} 10 \\ 50 \end{array}\right), \quad \mathbf{B}=\left(\begin{array}{r} 0 \\ -20 \end{array}\right) \end{aligned} $$

Step-by-Step Solution

Answer

1. \(x_1 = \frac{1}{4}, x_2 = -\frac{13}{8}\); 2. \(x_1 = -10, x_2 = -40\); 3. \(x_1 = 5, x_2 = 17.5\).

Step 1: Write the System in Matrix Form

Given the equations \(7x_1 - 2x_2 = b_1\) and \(3x_1 - 2x_2 = b_2\), express these in matrix form as \(\mathbf{AX} = \mathbf{B}\). Here \(\mathbf{A} = \begin{pmatrix} 7 & -2 \ 3 & -2 \end{pmatrix}\) and \(\mathbf{X} = \begin{pmatrix} x_1 \ x_2 \end{pmatrix}\).

Step 2: Calculate the Inverse of Matrix A

Calculate the inverse \(\mathbf{A}^{-1}\) of matrix \(\mathbf{A}\) by using the formula for a 2x2 matrix: \(\mathbf{A}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\). For \(\mathbf{A} = \begin{pmatrix} 7 & -2 \ 3 & -2 \end{pmatrix}\), we have \(ad-bc = (7)(-2) - (-2)(3) = -14 + 6 = -8\). Thus, \(\mathbf{A}^{-1} = \frac{1}{-8}\begin{pmatrix} -2 & 2 \ -3 & 7 \end{pmatrix}\).

Step 3: Solve for X using First Matrix B

Substitute \(\mathbf{B} = \begin{pmatrix} 5 \ 4 \end{pmatrix}\) into the equation \(\mathbf{X} = \mathbf{A}^{-1} \mathbf{B}\). Calculate \(\mathbf{X} = \frac{1}{-8}\begin{pmatrix} -2 & 2 \ -3 & 7 \end{pmatrix}\begin{pmatrix} 5 \ 4 \end{pmatrix} = \frac{1}{-8}\begin{pmatrix} (-2)(5) + (2)(4) \ (-3)(5) + (7)(4) \end{pmatrix} = \frac{1}{-8}\begin{pmatrix} -2 \ 13 \end{pmatrix}\). Thus, \(\mathbf{X} = \begin{pmatrix} \frac{1}{4} \ -\frac{13}{8} \end{pmatrix}\).

Step 4: Solve for X using Second Matrix B

Substitute \(\mathbf{B} = \begin{pmatrix} 10 \ 50 \end{pmatrix}\) into \(\mathbf{X} = \mathbf{A}^{-1} \mathbf{B}\). Calculate \(\mathbf{X} = \frac{1}{-8}\begin{pmatrix} -2 & 2 \ -3 & 7 \end{pmatrix}\begin{pmatrix} 10 \ 50 \end{pmatrix} = \frac{1}{-8}\begin{pmatrix} (-2)(10) + (2)(50) \ (-3)(10) + (7)(50) \end{pmatrix} = \frac{1}{-8}\begin{pmatrix} 80 \ 320 \end{pmatrix}\). Thus, \(\mathbf{X} = \begin{pmatrix} -10 \ -40 \end{pmatrix}\).

Step 5: Solve for X using Third Matrix B

Substitute \(\mathbf{B} = \begin{pmatrix} 0 \ -20 \end{pmatrix}\) into \(\mathbf{X} = \mathbf{A}^{-1} \mathbf{B}\). Calculate \(\mathbf{X} = \frac{1}{-8}\begin{pmatrix} -2 & 2 \ -3 & 7 \end{pmatrix}\begin{pmatrix} 0 \ -20 \end{pmatrix} = \frac{1}{-8}\begin{pmatrix} (-2)(0) + (2)(-20) \ (-3)(0) + (7)(-20) \end{pmatrix} = \frac{1}{-8}\begin{pmatrix} -40 \ -140 \end{pmatrix}\). Thus, \(\mathbf{X} = \begin{pmatrix} 5 \ 17.5 \end{pmatrix}\).

Key Concepts

Matrix Representation

When tackling a system of linear equations, one of the most effective ways to organize and solve such problems is through matrix representation. This involves expressing the system in the form \( \mathbf{A} \mathbf{X} = \mathbf{B} \).
- **System of equations**: In our context, you have equations like \( 7x_1 - 2x_2 = b_1 \) and \( 3x_1 - 2x_2 = b_2 \).
- **Matrix A**: represents the coefficients of the variables \( x_1 \) and \( x_2 \) in a compact form as \( \mathbf{A} = \begin{pmatrix} 7 & -2 \ 3 & -2 \end{pmatrix} \).
- **Matrix X**: is the column matrix of the variables \( x_1 \) and \( x_2 \), expressed as \( \mathbf{X} = \begin{pmatrix} x_1 \ x_2 \end{pmatrix} \).
- **Matrix B**: contains the results of the equations \( b_1, b_2 \) in a column vector form like \( \mathbf{B} = \begin{pmatrix} 5 \ 4 \end{pmatrix} \).
This form offers a clear route to leveraging matrix operations to find solutions.

Matrix Inversion

Matrix inversion is a crucial concept when working with systems of linear equations, particularly when the matrix \( \mathbf{A} \) is invertible.
- **Invertible Matrix**: For a matrix \( \mathbf{A} \), the inverse \( \mathbf{A}^{-1} \) exists if \( \mathbf{A} \) is square (same number of rows and columns) and has a non-zero determinant.
- **Formula**: For a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the inverse is calculated using \( \mathbf{A}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \).
- **Example Application**: For \( \mathbf{A} = \begin{pmatrix} 7 & -2 \ 3 & -2 \end{pmatrix} \), its determinant is \(-8\), and the inverse becomes \( \mathbf{A}^{-1} = \frac{1}{-8}\begin{pmatrix} -2 & 2 \ -3 & 7 \end{pmatrix} \).
Inverting the matrix \( \mathbf{A} \) allows us to solve the system \( \mathbf{A} \mathbf{X} = \mathbf{B} \) by finding \( \mathbf{X} = \mathbf{A}^{-1} \mathbf{B} \).

Solving Linear Systems

To solve a system of linear equations, matrix inversion plays a crucial role. By determining \( \mathbf{X} = \mathbf{A}^{-1} \mathbf{B} \), we can solve for the variables \( x_1 \) and \( x_2 \).
- **Step-by-Step Process**: Substitute each matrix \( \mathbf{B} \) into the equation \( \mathbf{X} = \mathbf{A}^{-1} \mathbf{B} \).
- **Example Calculation**: For \( \mathbf{B} = \begin{pmatrix} 5 \ 4 \end{pmatrix} \), compute \( \mathbf{X} = \frac{1}{-8}\begin{pmatrix} -2 & 2 \ -3 & 7 \end{pmatrix}\begin{pmatrix} 5 \ 4 \end{pmatrix} \), which results in \( \mathbf{X} = \begin{pmatrix} \frac{1}{4} \ -\frac{13}{8} \end{pmatrix} \).
This method can be applied to different \( \mathbf{B} \) values, giving solutions for each scenario. This process highlights the power of linear algebra in solving complex systems efficiently.

2x2 Matrices

2x2 matrices are one of the simplest forms of matrices yet they are incredibly useful in representing systems of linear equations with two variables.

• **Simplicity**: They are small but mighty, encapsulating essential characteristics of linear systems.
• **Multiplication**: To multiply a 2x2 matrix by another 2x1 matrix (like \( \mathbf{X} \)), each element of the resulting matrix is the sum of products of rows and corresponding columns.
• **Determinant**: The determinant of a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is \( ad - bc \), crucial for understanding invertibility.

By grasping the properties and operations of 2x2 matrices, one gains foundational insight into more complex linear algebra and systems.