Problem 21 - A Graphical Approach to College Algebra

Question

Each augmented matrix is in row echelon form and represents a linear system. Use back-substitution to solve the system if possible. $$\left[\begin{array}{rr|r} 1 & 2 & 3 \\ 0 & 1 & -1 \end{array}\right]$$

Step-by-Step Solution

Answer

The solution is \( x = 5 \) and \( y = -1 \).

Step 1: Understanding the Augmented Matrix

The given augmented matrix is \( \left[\begin{array}{rr|r} 1 & 2 & 3 \ 0 & 1 & -1 \end{array}\right] \). It represents a system of linear equations where the first row corresponds to the equation \( x + 2y = 3 \) and the second row corresponds to \( y = -1 \).

Step 2: Identify Equations from Matrix

From the second row of the augmented matrix, we directly get the equation \( y = -1 \). This solution for \( y \) can be immediately used for back-substitution.

Step 3: Back-Substitution for First Equation

Using the value of \( y = -1 \), substitute it into the first equation from the matrix: \( x + 2(-1) = 3 \). Simplify this to \( x - 2 = 3 \).

Step 4: Solve for x

Solve the equation \( x - 2 = 3 \) by adding 2 to both sides to isolate \( x \). Thus, \( x = 5 \).

Step 5: Verify the Solution

Substitute \( x = 5 \) and \( y = -1 \) back into the original equations to ensure they satisfy both. For the first equation: \( 5 + 2(-1) = 3 \), which is true, confirming our solution is correct.

Key Concepts

Augmented Matrix

An augmented matrix is a compact and efficient way to represent a system of linear equations. It combines the coefficient matrix of the variables and the constant terms from the equations into one matrix form. For example, given the system of equations:

• Equation 1: \(x + 2y = 3\)
• Equation 2: \(y = -1\)

the augmented matrix representation is:\[\left[\begin{array}{rr|r} 1 & 2 & 3 \ 0 & 1 & -1 \end{array}\right]\]Here, the vertical line separates the coefficients of the variables from the constant terms, assisting in visualizing and solving the system with techniques like Gaussian elimination or back-substitution.

Row Echelon Form

Row echelon form (REF) is a simplified version of a matrix that makes solving linear systems easier. A matrix is in row echelon form if:

• All nonzero rows are above any rows of all zeros.
• The leading coefficient (first nonzero number from the left, also known colloquially as the 'pivot') in a nonzero row is to the right of the leading coefficient of the row above it.
• All entries in a column below a leading entry are zeros.

Our example matrix:\[\left[\begin{array}{rr|r} 1 & 2 & 3 \ 0 & 1 & -1 \end{array}\right]\]shows how convenient this form is. It directly allows us to see solutions or use back-substitution, as the matrix steps down from left to right with zeros appearing below pivots.

Linear System

A linear system is a collection of one or more linear equations involving the same set of variables. In mathematical terms, it's a group of equations that can be written in the form:

• \(a_1x_1 + a_2x_2 + \cdots + a_nx_n = b\)

where \(a_1, a_2, ..., a_n\) are constants, \(x_1, x_2, ..., x_n\) are variables, and \(b\) is a constant term.In our exercise, we have a linear system:

• \(x + 2y = 3\)
• \(y = -1\)

Linear systems can have a single solution, infinitely many solutions, or no solution at all, depending on how the equations intersect or overlap. Representing them in an augmented matrix and reducing to row echelon form helps us identify which scenario we are dealing with.

Solving Equations

Solving equations is the process of finding the values for the variables that satisfy all equations in a system. The augmented matrix and row echelon form are tools that simplify this process. For our system:

• First, identify the easy-to-solve equation from the augmented matrix (in our case, \(y = -1\)).
• Use back-substitution to find other variable values. Start from the simplest equation known (usually the last row of the row echelon form) and work upwards.
• For the equation \(x + 2y = 3\), substitute \(y = -1\) in and solve for \(x\).
• Simplify the equation to find \(x = 5\).
Finally, substitute back into the original equations to verify the solution satisfies all conditions.