Problem 25 - A Graphical Approach to Precalculus with Limits

Question

Each augmented matrix is in row echelon form and represents a linear system. Use back-substitution to solve the system if possible. $$\left[\begin{array}{rrr|r} 1 & 1 & -1 & 4 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 1 & 1 \end{array}\right]$$

Step-by-Step Solution

Answer

The solution is \(x = 2\), \(y = 3\), \(z = 1\).

Step 1: Understand the Augmented Matrix

The given augmented matrix \(\left[\begin{array}{rrr|r} 1 & 1 & -1 & 4 \ 0 & 1 & -1 & 2 \ 0 & 0 & 1 & 1 \end{array}\right]\) represents the coefficients of a system of linear equations in row echelon form. Each row corresponds to an equation.

Step 2: Write the System of Equations

Translate the augmented matrix into a system of linear equations:1. \(x + y - z = 4\)2. \(y - z = 2\)3. \(z = 1\)

Step 3: Solve for z using Back-Substitution

Start with the last equation, which is already solved:\(z = 1\).

Step 4: Substitute the Value of z into the Second Equation

Substitute \(z = 1\) into the second equation:\(y - 1 = 2\).Solve for \(y\) by adding 1 to both sides:\(y = 3\).

Step 5: Substitute the Values of y and z into the First Equation

Substitute \(y = 3\) and \(z = 1\) into the first equation:\(x + 3 - 1 = 4\).Simplify and solve for \(x\):\(x + 2 = 4\).Subtract 2 from both sides: \(x = 2\).

Step 6: Write the Solution to the System

The solution to the system of equations is \(x = 2\), \(y = 3\), \(z = 1\). This represents the point where all three planes intersect in three-dimensional space.

Key Concepts

Augmented Matrix

An augmented matrix is a powerful tool for solving systems of linear equations. It combines both the coefficients of the variables and the constants from the equations into one matrix. In this format, each row of the matrix corresponds to one equation, and each column corresponds to the coefficients of the variables or the constants.

In the given example, the augmented matrix is:

• \([1, 1, -1 | 4]\)
• \([0, 1, -1 | 2]\)
• \([0, 0, 1 | 1]\)

This matrix represents the system of equations. The last column, usually separated by a vertical line, corresponds to the constants on the right side of the equations. The rest of the matrix represents the coefficients of the variables \(x\), \(y\), and \(z\).

By observing the augmented matrix, we can interpret how the system of equations is structured, helping us visually simplify and solve it.

Row Echelon Form

Row echelon form (REF) is a simplified version of a matrix that makes it easier to solve systems of linear equations. In this form, the matrix has zeros below each leading entry in a stair-like pattern.

• Leading coefficient of each non-zero row is 1.
• Each leading 1 is to the right of the leading 1 in the row above.
• Rows with all zero elements are at the bottom of the matrix.

The matrix is said to be in row echelon form if it adheres to these criteria.For the matrix in this exercise:

• First row: \([1, 1, -1 | 4]\) - It has a leading 1 in the first column.
• Second row: \([0, 1, -1 | 2]\) - Leading 1 in the second column and a zero above.
• Third row: \([0, 0, 1 | 1]\) - The leading 1 is in the third column with zeros above.

The row echelon form is essential before performing back-substitution because it clearly indicates the steps needed to find each variable, beginning from the bottom row.

System of Linear Equations

A system of linear equations consists of multiple linear equations that share the same set of variables. Solving such systems means finding the values of these variables that satisfy all equations in the system simultaneously.

The example matrix represents the following system of equations:

• \(x + y - z = 4\)
• \(y - z = 2\)
• \(z = 1\)

Each equation corresponds to a plane in three-dimensional space, and the solution is the point where all these planes intersect. Solving systems of linear equations can involve various methods, from graphical representations to algebraic manipulations like matrix transformations or substitution.

Solution of Equations

The solution to a system of equations is the set of values for the variables that make all the equations true simultaneously. In our matrix that represents the system, using back-substitution:

• We start with the last equation, already solved, \(z = 1\).
• Substitute \(z\) into the second equation to solve for \(y\): \(y - 1 = 2\), which simplifies to \(y = 3\).
• Finally, substitute \(y\) and \(z\) into the first equation to solve for \(x\): \(x + 3 - 1 = 4\), simplifying to \(x = 2\).

Thus, the solution is \(x = 2\), \(y = 3\), and \(z = 1\). This solution represents a single point where the three planes intersect in three-dimensional space, confirming a unique solution to this system.