Chapter 8: Problem 51
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 \(\mathbf{B}\). $$ \begin{aligned} &7 x_{1}-2 x_{2}=b_{1} \\ &3 x_{1}-2 x_{2}=b_{2} \\ &\mathbf{B}=\left(\begin{array}{l} 5 \\ 4 \end{array}\right), \quad \mathbf{B}=\left(\begin{array}{c} 10 \\ 50 \end{array}\right), \quad \mathbf{B}=\left(\begin{array}{r} 0 \\ -20 \end{array}\right) \end{aligned} $$
Short Answer
Step by step solution
Write the System in Matrix Form
Calculate the Inverse of Matrix A
Solve for X using First Matrix B
Solve for X using Second Matrix B
Solve for X using Third Matrix B
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Representation
- **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
- **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
- **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
- **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.