Problem 13 - College Algebra with Modeling and Visualization

Question

The augmented matrix is in row-echelon form and represents a linear system. Solve the system by using backwand substitution, if possible. Write the solution as either an ordered pair or an ordered triple. $$ \left[\begin{array}{rr|r} 1 & 2 & 3 \\ 0 & 1 & -1 \end{array}\right] $$

Step-by-Step Solution

Answer

(5, -1)

Step 1: Identify the structure of the system

The given augmented matrix \( \begin{bmatrix} 1 & 2 & | & 3 \ 0 & 1 & | & -1 \end{bmatrix} \) corresponds to a system of equations:- \( x + 2y = 3 \)- \( y = -1 \)

Step 2: Solve for the variable in the last row

Since the system is already in row-echelon form, begin with the last row, which gives the solution for \( y \):\[y = -1\]

Step 3: Substitute back to find the other variable(s)

Substitute the value of \( y = -1 \) into the first equation:\[ x + 2(-1) = 3 \]Simplify and solve for \( x \):\[x - 2 = 3 \x = 5\]

Step 4: Write the solution as an ordered pair

The solutions for the variables are \( x = 5 \) and \( y = -1 \). Therefore, the solution to the system is the ordered pair \( (5, -1) \).

Key Concepts

Augmented Matrix

An augmented matrix is a powerful tool when solving systems of linear equations. It allows us to work with the coefficients of the variables and the constants in a compact format. Think of it as a simplified way to organize the information from a set of linear equations. For example, a system of equations like:

• \( x + 2y = 3 \)
• \( y = -1 \)

can be represented by the augmented matrix:\[\left[\begin{array}{rr|r}1 & 2 & 3 \0 & 1 & -1 \\end{array}\right]\]Here, each row corresponds to an equation, and each column (before the vertical line) corresponds to a variable's coefficient, while the section after the vertical line represents the constants on the other side of the equations. This format helps visually organize the equations and supports the application of various solution techniques like Gaussian elimination.

Row-Echelon Form

The process of transforming a matrix to row-echelon form is essential in solving linear systems through methods such as Gaussian elimination. A matrix is in row-echelon form when:

• All rows consisting entirely of zeros are at the bottom.
• The leading entry (or pivot) of each nonzero row is 1.
• The leading entry (pivot) of a row is always to the right of the leading entry of the row above it.

In the given augmented matrix:\[\left[\begin{array}{rr|r}1 & 2 & 3 \0 & 1 & -1 \\end{array}\right]\]we can see that it is already in row-echelon form. The first leading 1 is in the first column, first row, and the next leading 1 is further to the right in the second row. This form simplifies solving systems by allowing straightforward application of backward substitution, leading us directly to the solution.

Backward Substitution

Backward substitution is a strategy used to solve systems of equations that results from row-echelon form. It involves starting from the bottom equation and working upwards to determine the values of each variable. Here’s how it works for the given matrix:1. **Start with the last row**: From the matrix: \[ 0y + 1y = -1 \] easily find that \( y = -1 \).2. **Substitute known values into previous equations**: Plug \( y = -1 \) into the first equation: \( x + 2(-1) = 3 \) which simplifies to: \( x = 5 \).3. **Conclude with an ordered pair**: Once the values are found, summarize the solution: \( (5, -1) \).This method efficiently solves systems of equations by transforming a row-echelon form matrix into its explicit solutions, aligning perfectly with sequential numerical logic.