Problem 81 - Precalculus with Limits

Question

In Exercises 63-84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. \( \left\\{ \begin{array}{l} -x + y = -22 \\ 3x + 4y = 4 \\ 4x - 8y = 32 \end{array} \right. \)

Step-by-Step Solution

Answer

The given system of equations has no solution.

Step 1: Write the System in Matrix form

As a first step, write the given system of linear equations in matrix form. This gives the augmented matrix: \[ \begin{bmatrix} -1 & 1 & -22 \ 3 & 4 & 4 \ 4 & -8 & 32 \end{bmatrix} \]

Step 2: Apply Gaussian elimination

To begin the Gaussian elimination, make the first entry of the first row positive by multiplying the first row by -1. After that swap the first row and the third row to get a matrix with leading entries as 1. This yields: \[ \begin{bmatrix} 4 & -8 & 32 \ 3 & 4 & 4 \ 1 & -1 & 22 \end{bmatrix} \]. Next, subtract 3/4 times the third row from the second row and 4 times the third row from the first row for the row echelon form. This gives \[ \begin{bmatrix} 0 & 4 & -56 \ 0 & 5 & -12.5 \ 1 & -1 & 22 \end{bmatrix} \]. Lastly, add the second row to the first to achieve the upper triangular form: \[ \begin{bmatrix} 0 & 9 & -68.5 \ 0 & 5 & -12.5 \ 1 & -1 & 22 \end{bmatrix} \]

Step 3: Solve for the variables using back substitution

From the last row, we have \(x = 22 + y\). From the second row, we have \(5y = -12.5\) which gives \(y = -2.5\). Substituting \(y = -2.5\) in \(x = 22 + y\) we get \(x = 19.5\). For the first row, we have \(9y = -68.5\) which gives \(y = -7.61\) to the 2 d.p. But y should be consistent throughout, hence there is no solution to this system of equations.

Key Concepts

System of Equations

A system of equations is like a puzzle we want to solve together. We find multiple equations with several variables. The solution gives us a set of values for these variables, making each equation true. Understanding how to solve such systems is essential in mathematics.
In the given exercise, we have a set of three equations with variables x and y:

• \(-x + y = -22\)
• \(3x + 4y = 4\)
• \(4x - 8y = 32\)

These equations are interconnected. Finding a solution means identifying values for x and y that satisfy all three equations. Sometimes, as in this exercise, due to inconsistency, no solution exists.

Matrix Form

Matrices are like friendly tables that help organize the numbers and variables of equations. When solving a system of equations, turning them into matrix form is the first step.
A matrix displays coefficients and constants from each equation systematically. For the exercise, the matrix form is given as:\[\begin{bmatrix}-1 & 1 & -22 \3 & 4 & 4 \4 & -8 & 32 \\end{bmatrix}\]This matrix has three rows, each representing an equation, and columns for each variable and constants.
In matrix form, solving the problem becomes more straightforward as it enables us to use methods like Gaussian elimination.

Back Substitution

Back substitution is a process we use after transforming our matrix into an upper triangular form.
Starting from the bottom row, it helps us "substitute back" known values to find unknowns in other equations. Here, the goal is to work backward from the last row to the first.
In the given steps, we used back substitution to find the values of x and y. Here's how it worked:

• The last row provided \(x = 22 + y\).
• The second row gave \(5y = -12.5\), which we solved to find \(y = -2.5\).
• Substituting \(y = -2.5\) into \(x = 22 + y\) yielded \(x = 19.5\).
• Lastly, there was a contradiction in the first row's result, showing no consistent solution exists.

Back substitution is invaluable in confirming solutions' consistency.

Gauss-Jordan Elimination

Gauss-Jordan elimination is an advanced matrix technique used to simplify and solve system equations thoroughly. While similar to Gaussian elimination, it goes further to ensure a matrix's diagonal consists entirely of ones (and zeros elsewhere).
To perform Gauss-Jordan elimination, we:

• Transform the matrix into a row-echelon form where leading coefficients are 1.
• Eliminate coefficients above and below leading ones to simplify further.

In our case, though both Gaussian elimination and back substitution were applied, issues arose. Ultimately using Gauss-Jordan elimination would've helped clarify no solutions existed due to inconsistent results. Implementing Gauss-Jordan fully, then checking for contradictions in the variables ensures more accurate problem resolution.