Problem 22 - 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. $$\left[\begin{array}{rr|r} 1 & -1 & 2 \\ 0 & 1 & 0 \end{array}\right]$$

Step-by-Step Solution

Answer

The solution to the system is \(x = 2\) and \(y = 0\).

Step 1: Understand the Row Echelon Form

The given augmented matrix is \( \begin{bmatrix} 1 & -1 & | & 2 \ 0 & 1 & | & 0 \end{bmatrix} \), which corresponds to the linear system: \(\begin{align*} x - y &= 2 \ y &= 0 \end{align*}\). Our task is to find the values of \(x\) and \(y\) using back-substitution.

Step 2: Substitute the Known Value

From the second row of the matrix, we directly have \(y = 0\). This is our known value, and we will substitute it into the first equation to find \(x\).

Step 3: Solve for the Unknown Value

Substitute \(y = 0\) into the first equation, \(x - y = 2\). This simplifies to \(x - 0 = 2\) or \(x = 2\).

Step 4: Confirm the Solution

Now, we have \(y = 0\) and \(x = 2\). These values satisfy both equations of the system, confirming that the solution is correct.

Key Concepts

Row Echelon Form

When studying linear systems, transforming the system's equations into a row echelon form (REF) can simplify solving them. This form is characterized by a staircase pattern in the coefficients of the variables, achieved through row operations:

• Each leading entry of a row (called a pivot) is to the right of the pivot of the row above it.
• All entries in a column below a pivot are zeros.
• Any rows consisting entirely of zeros are at the bottom of the matrix.

In the given exercise, the matrix \[\left[\begin{array}{rr|r} 1 & -1 & 2 \0 & 1 & 0 \end{array}\right] \] is already in row echelon form. The first row corresponds to the equation \(x - y = 2\), and the second row provides \(y = 0\). This structured format facilitates the back-substitution process to find the solutions easily.

Linear System

A linear system is a collection of linear equations that involves the same set of variables. Solving a linear system means finding the values of these variables that satisfy all the equations simultaneously. Linear systems can have:

• Exactly one solution (consistent and independent).
• Infinitely many solutions (consistent and dependent).
• No solution (inconsistent).

In the given system, we deal with two variables, \(x\) and \(y\), represented by the equations \(x - y = 2\) and \(y = 0\). Since we have two variables and two equations, we're usually looking for a unique solution. After back-substitution, we determine \(y = 0\) and \(x = 2\), meaning the system is consistent with exactly one solution.

Augmented Matrix

An augmented matrix is a compact and efficient way to represent a linear system. It consists of the coefficient matrix of the system combined with an additional column representing the constants from the equations:

• The rows of the augmented matrix correspond to the equations of the linear system.
• The last column represents the constants from each equation.

For example, in the matrix:\[\left[\begin{array}{rr|r} 1 & -1 & 2 \0 & 1 & 0 \end{array}\right] \]the elements on the right-hand side of the vertical line form the constants of each equation, turning it into an easy visual format. This arrangement simplifies the process of applying row operations such as finding solutions through back-substitution, making understanding and solving linear systems more straightforward.