Problem 15 - Elementary and Intermediate Algebra

Question

Solve using Cramer's rule. \(5 x+8 y=1\) \(3 x+7 y=5\)

Step-by-Step Solution

Answer

The solution is \(x = -3\) and \(y = 2\).

Step 1: Write the system of equations in matrix form

The given system of equations is: \[5x + 8y = 1\]\[3x + 7y = 5\]This can be written in matrix form as \(AX = B\), where \[A = \begin{pmatrix} 5 & 8 \ 3 & 7 \end{pmatrix}, \ X = \begin{pmatrix} x \ y \end{pmatrix}, \ B = \begin{pmatrix} 1 \ 5 \end{pmatrix}\]

Step 2: Calculate the determinant of the coefficient matrix A

The determinant \(\Delta\) of matrix A is calculated as follows: \[\Delta = \begin{vmatrix} 5 & 8 \ 3 & 7 \end{vmatrix} = 5 \times 7 - 8 \times 3 = 35 - 24 = 11\]

Step 3: Form the matrix A_x and calculate its determinant

Replace the first column of matrix A with the constants matrix B to form \(A_x\): \[A_x = \begin{pmatrix} 1 & 8 \ 5 & 7 \end{pmatrix}\]The determinant \(\Delta_x\) of matrix \(A_x\) is: \[\Delta_x = \begin{vmatrix} 1 & 8 \ 5 & 7 \end{vmatrix} = 1 \times 7 - 8 \times 5 = 7 - 40 = -33\]

Step 4: Form the matrix A_y and calculate its determinant

Replace the second column of matrix A with the constants matrix B to form \(A_y\): \[A_y = \begin{pmatrix} 5 & 1 \ 3 & 5 \end{pmatrix}\]The determinant \(\Delta_y\) of matrix \(A_y\) is: \[\Delta_y = \begin{vmatrix} 5 & 1 \ 3 & 5 \end{vmatrix} = 5 \times 5 - 1 \times 3 = 25 - 3 = 22\]

Step 5: Apply Cramer's rule to solve for x and y

Using Cramer's rule, the solutions for \(x\) and \(y\) are given by: \[x = \frac{\Delta_x}{\Delta} = \frac{-33}{11} = -3\]\[y = \frac{\Delta_y}{\Delta} = \frac{22}{11} = 2\]

Key Concepts

Determinant of a Matrix

The determinant is a special number that can be calculated from a square matrix. It is crucial in various calculations in linear algebra, such as solving systems of linear equations using Cramer's rule. The determinant of a 2x2 matrix \([a \ b \ c \ d]\) is calculated using the formula: \[ \text{determinant} = ad - bc \] This unique value helps determine if a system has a unique solution, no solution, or infinitely many solutions. If the determinant is zero, the matrix is singular, meaning the system of equations does not have a unique solution. For our matrix A in the problem: \[ A = \begin{pmatrix} 5 \ 8 \ 3 \ 7 \end{pmatrix} \] The determinant (\text{det}(A)) is calculated as follows: \[ \text{det}(A) = 5 \times 7 - 8 \times 3 = 35 - 24 = 11 \] The non-zero determinant implies that the system of equations has a unique solution.

System of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. In our exercise, we have the system: \[ 5x + 8y = 1 \ 3x + 7y = 5 \] Each equation represents a line in a two-dimensional space. The solution to the system is the point where these lines intersect. If the lines intersect at a single point, there is a unique solution. If they are parallel, there is no solution. If the lines are identical, there are infinitely many solutions. The system can be expressed in matrix form as AX = B, where A is the coefficient matrix, X is the column matrix of variables x and y, and B is the constants matrix: \[ A = \begin{pmatrix} 5 \ 8 \ 3 \ 7 \end{pmatrix}, \ X = \begin{pmatrix} x \ y \end{pmatrix}, \ B = \begin{pmatrix} 1 \ 5 \end{pmatrix} \] This form is helpful for using matrix algebra techniques to find solutions.

Matrix Algebra

Matrix algebra involves the manipulation of matrices to solve systems of linear equations, among other applications. Key operations include calculating determinants, matrix multiplication, and finding inverses. When solving a system using Cramer's rule, we perform specific steps involving determinants of various matrices. First, we find the determinant of the original coefficient matrix A. If it's non-zero, we proceed to form matrices \(A_x\) and \(A_y\) by replacing columns of A with the constants matrix B. For matrix \(A_x\): \[ A_x = \begin{pmatrix} 1 \ 8 \ 5 \ 7 \end{pmatrix} \] The determinant of \(A_x\) is: \[ \text{det}(A_x) = 1 \times 7 - 8 \times 5 = 7 - 40 = -33 \] For matrix \(A_y\): \[ A_y = \begin{pmatrix} 5 \ 1 \ 3 \ 5 \end{pmatrix} \] The determinant of \(A_y\) is: \[ \text{det}(A_y) = 5 - 1 \times 3 = 25 - 3 = 22 \] Finally, we apply Cramer's rule: \[ x = \frac{\Delta_x}{\Delta} = \frac{-33}{11} = -3 \] \[ y = \frac{\Delta_y}{\Delta} = \frac{22}{11} = 2 \] Matrix algebra simplifies such complex calculations and makes it easier to find solutions to linear systems.