Chapter 9: Problem 33
Solve using Cramer's rule. $$\begin{aligned}&2 x+9 y=-2\\\&4 x-3 y=3\end{aligned}$$
Short Answer
Expert verified
The solution is \(x = \frac{1}{2}\) and \(y = -\frac{1}{3}\).
Step by step solution
01
- Write the system in matrix form
Express the system of equations as a matrix equation: \[ A \mathbf{x} = \mathbf{b} \] where matrix \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variable matrix, and \( \mathbf{b} \) is the constant matrix. For the given system, we have: \[ A = \begin{pmatrix} 2 & 9 \ 4 & -3 \end{pmatrix}, \mathbf{x} = \begin{pmatrix} x \ y \end{pmatrix}, \mathbf{b} = \begin{pmatrix} -2 \ 3 \end{pmatrix} \]
02
- Find the determinant of matrix A
Calculate the determinant of matrix \( A \): \[ \text{det}(A) = \begin{vmatrix} 2 & 9 \ 4 & -3 \end{vmatrix} = (2 \times (-3)) - (9 \times 4) = -6 - 36 = -42 \]
03
- Find the determinant of matrix A with respect to x (\( A_x \))
Replace the first column of \( A \) with vector \( \mathbf{b} \) and find the determinant: \[ A_x = \begin{pmatrix} -2 & 9 \ 3 & -3 \end{pmatrix} \text{det}(A_x) = (-2 \times (-3)) - (9 \times 3) = 6 - 27 = -21 \]
04
- Find the determinant of matrix A with respect to y (\( A_y \))
Replace the second column of \( A \) with vector \( \mathbf{b} \) and find the determinant: \[ A_y = \begin{pmatrix} 2 & -2 \ 4 & 3 \end{pmatrix} \text{det}(A_y) = (2 \times 3) - ((-2) \times 4) = 6 + 8 = 14 \]
05
- Solve for x and y
Using Cramer's rule: \[ x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{-21}{-42} = \frac{1}{2} \] \[ y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{14}{-42} = -\frac{1}{3} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
determinants
Determinants are a special number calculated from a square matrix. For a 2x2 matrix, the determinant is computed as follows: \ For matrix \ \ \[ A = \begin{pmatrix} a & b \ c & d \ \end{pmatrix} \] \ \ The determinant of matrix \( A \) is denoted \( \text{det}(A) \) or \( |A| \) and is given by: \ \ \[ \text{det}(A) = ad - bc \] \ In our exercise, matrix \( A \) was \ \[ \begin{pmatrix} 2 & 9 \ 4 & -3 \end{pmatrix} \] \ By formula, we computed: \ \[ \text{det}(A) = (2 \cdot (-3)) - (9 \cdot 4) = -6 - 36 = -42 \] \ This determinant plays a crucial role in solving systems of linear equations using Cramer’s Rule. Without a non-zero determinant, Cramer's Rule cannot be applied.
systems of linear equations
A system of linear equations is a collection of one or more linear equations involving the same set of variables. Here, we had the following system: \ \ \[ \begin{aligned} 2x + 9y &= -2 \ 4x - 3y &= 3 \end{aligned} \] \ The goal is to find the values of \( x \) and \( y \) that satisfy both equations simultaneously. This type of problem can be effectively solved using matrix methods like Cramer's Rule, especially because these methods clearly organize the coefficients and constants in a structured format.
matrix algebra
Matrix algebra is a branch of algebra that deals with matrices and operations such as addition, subtraction, and multiplication of matrices. In this exercise, matrix algebra helped to reformulate the system of equations in a compact matrix form. \ \ The given system of equations was converted to: \ \ \[ A \mathbf{x} = \mathbf{b} \] \ \ Here, \ \ \[ A = \begin{pmatrix} 2 & 9 \ 4 & -3 \end{pmatrix}, \mathbf{x} = \begin{pmatrix} x \ y \ \end{pmatrix}, \mathbf{b} = \begin{pmatrix} -2 \ 3 \end{pmatrix} \] \ \ Each element in the matrix \( A \) represents a coefficient from the system of linear equations, while \( \mathbf{x} \) and \( \mathbf{b} \) consist of the variables and constants, respectively. This matrix formulation allows for the utilization of powerful matrix operations and rules such as Cramer’s Rule to find the solution efficiently.