Problem 64 - Intermediate Algebra

Question

Use Cramer's rule to solve each system of equations. If a system is inconsistent or if the equations are dependent, so indicate. $$ \left\\{\begin{array}{l} 2 x+5 y-13=0 \\ -2 x+13=5 y \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = 0 \) and \( y = 2.6 \). The system is consistent and independent.

Step 1: Write the system in standard form

First, express each equation in the form \( ax + by = c \). The given system is already close to standard form:\[ \begin{align*} 2x + 5y &= 13 \ -2x + 5y &= 13 \end{align*} \]

Step 2: Identify the coefficients

From the equations, the coefficients are:- For the first equation: \( a_1 = 2, b_1 = 5, c_1 = 13 \).- For the second equation: \( a_2 = -2, b_2 = 5, c_2 = 13 \).

Step 3: Write the coefficient matrix and constants

The coefficient matrix \( A \) and the constant matrix \( B \) are:\[ A = \begin{bmatrix} 2 & 5 \ -2 & 5 \end{bmatrix} \]\[ B = \begin{bmatrix} 13 \ 13 \end{bmatrix} \]

Step 4: Calculate the determinant of the coefficient matrix

Compute the determinant \( \Delta \) of matrix \( A \):\[ \Delta = \begin{vmatrix} 2 & 5 \ -2 & 5 \end{vmatrix} = (2)(5) - (5)(-2) = 10 + 10 = 20 \]

Step 5: Calculate determinant for x

Replace the first column of \( A \) with the constants from matrix \( B \) and calculate determinant \( \Delta_x \):\[ \Delta_x = \begin{vmatrix} 13 & 5 \ 13 & 5 \end{vmatrix} = (13)(5) - (5)(13) = 65 - 65 = 0 \]

Step 6: Calculate determinant for y

Replace the second column of \( A \) with the constants and calculate determinant \( \Delta_y \):\[ \Delta_y = \begin{vmatrix} 2 & 13 \ -2 & 13 \end{vmatrix} = (2)(13) - (-2)(13) = 26 + 26 = 52 \]

Step 7: Determine solutions using Cramer's rule

Using Cramer's rule, find \( x \) and \( y \):\[ x = \frac{\Delta_x}{\Delta} = \frac{0}{20} = 0 \]\[ y = \frac{\Delta_y}{\Delta} = \frac{52}{20} = 2.6 \]

Step 8: Check for consistency

Both the determinant \( \Delta \) and the solution for \( x \) and \( y \) are valid because \( \Delta eq 0 \). The system is consistent and the equations are independent.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with a set of unknowns. In our exercise, we have two linear equations with two variables, \( x \) and \( y \). Our aim is to find values for these variables that satisfy both equations simultaneously. When solving these systems, the equations can either intersect at a single point (one solution), be parallel with no intersection (no solution), or be identical thus intersecting at infinitely many points (infinitely many solutions). In this particular problem, we are tasked with finding a single solution using Cramer's Rule, a technique that involves matrix calculations and determinants to provide a clear, algebraic solution to such problems.

Determinants

The determinant is a special number that can be calculated from a square matrix. It provides important information about the matrix and the system of equations it represents. Calculating the determinant is crucial when applying Cramer's Rule because it allows you to determine whether a unique solution exists. In simple terms, if the determinant is non-zero, the system has a unique solution; if zero, the system may either have no solutions or an infinite number of solutions. In our solution, we first calculated the determinant of the coefficient matrix:

• For matrix \( A = \begin{bmatrix} 2 & 5 \ -2 & 5 \end{bmatrix} \), the determinant is \( \Delta = 20 \) which means a unique solution exists.
• We also calculated "partial determinants" for \( x \) and \( y \) needed for Cramer's Rule. These partial determinants help find the specific values for each variable.

Matrix Algebra

Matrix algebra is a branch of mathematics that deals with operations on matrices, which are rectangular arrays of numbers or variables. In the context of solving systems of equations, we use matrices to organize and process these systems more efficiently. In the exercise at hand, matrix algebra is employed to express the system's coefficients in a more structured form.

• The coefficient matrix \( A \) is \( \begin{bmatrix} 2 & 5 \ -2 & 5 \end{bmatrix} \). This matrix includes all coefficients of the system's variables without their constants.
• The constant matrix \( B \) is \( \begin{bmatrix} 13 \ 13 \end{bmatrix} \). This matrix lists the constants from the right-hand sides of the system equations.

Using these matrices allows us to easily compute determinants that lead to the solution for each variable. This structured approach is part of the beauty of matrix algebra, providing efficiency and clarity in handling complicated systems.