Problem 17 - 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. $$ \left[\begin{array}{rrr|r} 1 & 1 & -1 & 4 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 1 & 1 \end{array}\right] $$

Step-by-Step Solution

Answer

The solution is \((2, 3, 1)\).

Step 1: Identify the rows in the system

The augmented matrix represents the linear system of equations. From the matrix, we extract the equations: 1. The first row gives us: \( x + y - z = 4 \) 2. The second row gives us: \( y - z = 2 \) 3. The third row gives us: \( z = 1 \).

Step 2: Solve for the last variable

Since the third row yields the equation \( z = 1 \), we have determined that \( z = 1 \). This value will be used in the other equations.

Step 3: Substitute and solve for the second variable

Using the equation from the second row, \( y - z = 2 \), substitute \( z = 1 \) to find \( y \): \[y - 1 = 2 \] Adding 1 to both sides gives: \[y = 3 \].

Step 4: Substitute and solve for the first variable

Using the equation from the first row, \( x + y - z = 4 \), substitute the known values \( z = 1 \) and \( y = 3 \): \[x + 3 - 1 = 4 \] Simplify to \[x + 2 = 4 \] Subtract 2 from both sides to find \( x \): \[x = 2 \].

Step 5: Write the solution as an ordered triple

The solution to the linear system is the ordered triple consisting of the values for \( x \), \( y \), and \( z \): \((x, y, z) = (2, 3, 1)\).

Key Concepts

Row-Echelon Form

When working with matrices in linear algebra, one valuable form is the **row-echelon form**. This simple structure makes it easier to solve systems of linear equations. A matrix is in row-echelon form when:

• All nonzero rows are above any rows of all zeros.
• The leading coefficient (the first nonzero number from the left, also called a pivot) in each nonzero row is to the right of the leading coefficient of the row above it.
• Any zero rows are at the bottom of the matrix.

In the matrix given in the ORIGINAL EXERCISE:

• The first row has a leading 1, or pivot, in the first column.
• The second row has a leading 1 in the second column, and zero in the first column.
• The third row has a leading 1 in the third column, with zeros in the previous columns.

This staircase-like appearance is what makes the matrix easy to work with, as it clearly shows the steps needed to solve for each variable one at a time.

Back Substitution

Once a matrix is in row-echelon form, the next step is to use **back substitution** to find the solutions for the variables in the system of equations. Back substitution starts from the last equation and works its way up to the first.In the provided exercise, we have already established from the matrix that:

• The third row gives us the equation: \[z = 1\]We start by solving for \( z \).
• In the second row, knowing \( z = 1 \), we use the equation:\[y - z = 2\] We substitute \( z = 1 \) which allows us to solve for \( y \).
• In the first row, now with both \( y = 3 \) and \( z = 1 \), the equation:\[x + y - z = 4\]allows us to solve for \( x \).

The key here is moving upwards row by row, substituting the already known values to find the remaining unknowns. This method efficiently works through the equations, from those with the fewest unknowns to those with the most.

Linear Systems

In mathematics, a **linear system** refers to a collection of linear equations. Each equation in the system represents a line in the mathematical space. Solving a linear system means finding the values of the variables that satisfy all equations simultaneously.In the context of the given matrix:

• Each row translates directly to an equation with three variables \( x, y, \text{ and } z \).
• The goal is to find a point that lies on all three planes formed by these equations. In 3D space, this common point is represented as an ordered triple \((x, y, z)\).
• A solution exists when the equations intersect at a single point.

From the step-by-step guide, we gleaned the equations from the row-echelon form matrix and solved them through back substitution. This process showcased the direct application of linear algebra techniques to analyze and solve linear systems, providing a clear, ordered answer: \((2, 3, 1)\). This indicates the point where all the planes intersect, solving the system completely.