Problem 38 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 33-42: Use Cramer's rule, whenever applicable, to solve the system. $$ \left\\{\begin{array}{r} 3 p-q=7 \\ -12 p+4 q=3 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( p = \frac{31}{24} \) and \( q = \frac{31}{8} \).

Step 1: Confirm System's Suitability for Cramer's Rule

Cramer's Rule can be applied if the coefficient matrix of the system is square (same number of equations as unknowns) and has a non-zero determinant. Here, we have two equations and two unknowns, so the matrix is square.

Step 2: Write the Coefficient Matrix and Calculate Determinant

The coefficient matrix is \( A = \begin{pmatrix} 3 & -1 \ -12 & 4 \end{pmatrix} \). The determinant of \( A \) is calculated as: \[ \det(A) = (3)(4) - (-1)(-12) = 12 + 12 = 24 \].Since \( \det(A) eq 0 \), Cramer's Rule can be applied.

Step 3: Set Up Matrices for Cramer's Rule

To apply Cramer's Rule, calculate determinants of matrices formed by replacing columns of the coefficient matrix.- Matrix \( A_p \) is formed by replacing the first column of \( A \) with the constants: \( A_p = \begin{pmatrix} 7 & -1 \ 3 & 4 \end{pmatrix} \).- Matrix \( A_q \) is formed by replacing the second column of \( A \) with the constants: \( A_q = \begin{pmatrix} 3 & 7 \ -12 & 3 \end{pmatrix} \).

Step 4: Calculate Determinants of Modified Matrices

Calculate the determinant of \( A_p \): \[ \det(A_p) = (7)(4) - (-1)(3) = 28 + 3 = 31 \].Calculate the determinant of \( A_q \): \[ \det(A_q) = (3)(3) - (7)(-12) = 9 + 84 = 93 \].

Step 5: Solve for Variables Using Cramer's Rule

According to Cramer's Rule, the solutions for the unknowns are given by:\[ p = \frac{\det(A_p)}{\det(A)} = \frac{31}{24} \].\[ q = \frac{\det(A_q)}{\det(A)} = \frac{93}{24} = \frac{31}{8} \].Thus, \( p = \frac{31}{24} \) and \( q = \frac{31}{8} \).

Key Concepts

Determinant Calculation

The determinant of a matrix is a special number that can be calculated from its elements. It's essential when using Cramer's Rule as it determines if the rule can be applied. For a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the calculation of the determinant follows the formula:\[ \det(A) = ad - bc \]In our exercise, the coefficient matrix \( A \) is \( \begin{pmatrix} 3 & -1 \ -12 & 4 \end{pmatrix} \). Calculating its determinant involves substituting into the formula:\[ \det(A) = (3)(4) - (-1)(-12) = 12 + 12 = 24 \]Since the determinant \( \det(A) eq 0 \), this confirms that Cramer’s Rule is applicable.

Linear Equations

Linear equations represent relationships involving variables with constant coefficients and no exponents. They are typically written in the form \( ax + by = c \). In this exercise, we are working with a system of linear equations:- \( 3p - q = 7 \)- \( -12p + 4q = 3 \) Each equation in the system forms a straight line when graphed. The goal is to find the intersection of these lines, which corresponds to the solution of the system. Cramer's Rule provides a method to solve these equations, particularly when a unique solution exists, revealed by a non-zero determinant in the coefficient matrix.

Coefficient Matrix

The coefficient matrix is formed from the coefficients of the variables in the system of linear equations. It plays a pivotal role in solving the system using methods like Cramer's Rule. For our system:- The first equation is \( 3p - q = 7 \)- The second equation is \( -12p + 4q = 3 \)We derive the coefficient matrix \( A \) as:\[ A = \begin{pmatrix} 3 & -1 \ -12 & 4 \end{pmatrix} \]This matrix is square, meaning it has an equal number of rows and columns. For such matrices, a unique solution to the equation system is possible if the matrix's determinant is non-zero.

Determinant of Matrices

In the context of solving systems of linear equations, the determinant of a matrix serves as a crucial indicator. A non-zero determinant signifies that the system has a unique solution, while a zero determinant indicates no unique solution or infinitely many solutions.After forming the coefficient matrix \( A = \begin{pmatrix} 3 & -1 \ -12 & 4 \end{pmatrix} \), we determine the determinant to be\[ \det(A) = 24 \]Since \( \det(A) eq 0 \), we proceed with Cramer's Rule by constructing altered matrices, \( A_p \) and \( A_q \), replacing respective columns with the constants:- \( A_p = \begin{pmatrix} 7 & -1 \ 3 & 4 \end{pmatrix} \)- \( A_q = \begin{pmatrix} 3 & 7 \ -12 & 3 \end{pmatrix} \)We calculate their determinants:- \( \det(A_p) = 31 \)- \( \det(A_q) = 93 \)With these determinants, Cramer's Rule allows solving for the variables \( p \) and \( q \) in the equations.