Problem 43 - College Algebra

Question

For the following exercises, solve the system of linear equations using Cramer's Rule. $$ \begin{aligned} 4 x-6 y+8 z &=10 \\ -2 x+3 y-4 z &=-5 \\ x+y+z &=1 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \(x = 5\), \(y = 5\), \(z = 1\).

Step 1: Write down the coefficient matrix

The first step is to write the coefficient matrix \(A\) from the given system of equations:\[A = \begin{bmatrix} 4 & -6 & 8 \ -2 & 3 & -4 \ 1 & 1 & 1 \end{bmatrix}\]

Step 2: Calculate determinant of the coefficient matrix

Calculate the determinant of matrix \(A\), denoted as \(\det(A)\).Using the formula for the determinant of a 3x3 matrix:\[ \det(A) = 4(3 \times 1 - (-4) \times 1) - (-6)(-2 \times 1 - (-4) \times 1) + 8(-2 \times 1 - 3 \times 1)\]Evaluate this expression to find \(\det(A) = -6\).

Step 3: Calculate determinant for x (\(D_x\))

To find \(D_x\), replace the first column of the coefficient matrix \(A\) with the constants on the right side of the equations:\[D_x = \begin{vmatrix} 10 & -6 & 8 \ -5 & 3 & -4 \ 1 & 1 & 1 \end{vmatrix}\]Calculate the determinant:\[ D_x = 10(3 \times 1 - (-4) \times 1) - (-6)(-5 \times 1 - (-4) \times 1) + 8(-5 \times 1 - 3 \times 1)\]This evaluates to \(D_x = -30\).

Step 4: Calculate determinant for y (\(D_y\))

To find \(D_y\), replace the second column of \(A\) with the constants:\[D_y = \begin{vmatrix} 4 & 10 & 8 \ -2 & -5 & -4 \ 1 & 1 & 1 \end{vmatrix}\]Calculate the determinant:\[ D_y = 4(-5 \times 1 - (-4) \times 1) - 10(-2 \times 1 - (-4) \times 1) + 8(-2 \times 1 - (-5) \times 1)\]This evaluates to \(D_y = -30\).

Step 5: Calculate determinant for z (\(D_z\))

To find \(D_z\), replace the third column of \(A\) with the constants:\[D_z = \begin{vmatrix} 4 & -6 & 10 \ -2 & 3 & -5 \ 1 & 1 & 1 \end{vmatrix}\]Calculate the determinant:\[ D_z = 4(3 \times 1 - (-5) \times 1) - (-6)(-2 \times 1 - (-5) \times 1) + 10(-2 \times 1 - 3 \times 1)\]This evaluates to \(D_z = -6\).

Step 6: Apply Cramer's Rule

Using Cramer's Rule, solve for each variable:\[x = \frac{D_x}{\det(A)}, \quad y = \frac{D_y}{\det(A)}, \quad z = \frac{D_z}{\det(A)}\]Substitute the values:\[x = \frac{-30}{-6} = 5, \quad y = \frac{-30}{-6} = 5, \quad z = \frac{-6}{-6} = 1\]

Step 7: Write the solution

The solution to the system of equations is \(x = 5\), \(y = 5\), and \(z = 1\).

Key Concepts

Determinants

The concept of determinants is fundamental in linear algebra, especially when dealing with matrices. A determinant is a special value that is calculated from a square matrix. It can determine whether a system of linear equations has a unique solution, no solution, or infinitely many solutions. This is vital because it allows us to understand the solvability of systems of equations more profoundly.

When working with a 3x3 matrix, as in the exercise above, the determinant can be calculated by expanding along any row or column. This involves taking minor determinants and applying a checkerboard pattern of signs. The determinant of a matrix \(A\), denoted \(\det(A)\), is crucial for Cramer's Rule because it serves as the denominator when solving for variables. If \(\det(A)\) is zero, the system either has no solutions or infinitely many, which means Cramer's Rule cannot be applied.

Understanding how to compute the determinant is essential for applying Cramer's Rule effectively, as shown in the step-by-step solution. By determining \(\det(A) = -6\), we confirm the system has a unique solution, allowing us to proceed.

System of Equations

A system of equations consists of multiple equations working together. In the example provided, there are three equations with three variables: \(x\), \(y\), and \(z\). The goal is to find values for these variables that satisfy all equations at the same time. These kinds of problems frequently arise in real-world applications, such as physics, engineering, and economics, where multiple factors influence the outcome.

The system of equations is typically expressed as:

• \(4x - 6y + 8z = 10\)
• \(-2x + 3y - 4z = -5\)
• \(x + y + z = 1\)

In matrix form, this set is represented by a coefficient matrix, which simply organizes the coefficients of each variable into a matrix. Solving this system manually requires understanding how these equations interact and applying an appropriate method, like Cramer's Rule, to find a solution. The key here is setting up our matrices correctly while remembering that the solution is only valid if \(\det(A) eq 0\).

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, matrices, and systems of equations. It's incredibly powerful, offering a framework for many practical and theoretical exercises. At its core, linear algebra provides the language and tools necessary to describe linear systems, which are systems of equations where each equation is linear—meaning it graphs as a straight line in two dimensions, a plane in three dimensions, and so on.

In this exercise, linear algebra helps us understand how to manipulate equations into matrix form, which then allows us to employ Cramer's Rule. This method simplifies solving the system by using determinants, showcasing linear algebra's utility in transforming complex systems into more manageable numerical forms. As students delve deeper into linear algebra, they'll see its applications across various scientific disciplines, solidifying its status as a foundational mathematical tool.