Problem 30 - College Algebra with Modeling and Visualization

Question

Use Gaussian elimination with backward substitution to solve the system of linear equations. Write the solution as an ordered pair or an ordered triple whenever possible. $$ \begin{array}{r} x+y+z=6 \\ 2 x+3 y-z=3 \\ x+y+2 z=10 \end{array} $$

Step-by-Step Solution

Answer

The solution is \((-1, 3, 4)\).

Step 1: Write the Augmented Matrix

First, represent the system of equations in an augmented matrix form. The system can be written as follows:\[\begin{bmatrix} 1 & 1 & 1 & | & 6 \ 2 & 3 & -1 & | & 3 \ 1 & 1 & 2 & | & 10 \end{bmatrix}\]

Step 2: Perform Row Operations to Achieve Row Echelon Form

Our goal is to transform the matrix into an upper triangular form. We start with row operations:- Subtract the first row from the third row to eliminate the first position in the third row: \[ R_3 = R_3 - R_1 \rightarrow \begin{bmatrix} 1 & 1 & 1 & | & 6 \ 2 & 3 & -1 & | & 3 \ 0 & 0 & 1 & | & 4 \end{bmatrix} \]- Subtract 2 times the first row from the second row: \[ R_2 = R_2 - 2R_1 \rightarrow \begin{bmatrix} 1 & 1 & 1 & | & 6 \ 0 & 1 & -3 & | & -9 \ 0 & 0 & 1 & | & 4 \end{bmatrix} \]

Step 3: Perform Backward Substitution

Now that the matrix is in row echelon form, use backward substitution to find the values of \(x\), \(y\), and \(z\):- From the third row, \[ z = 4 \].- Substitute \( z = 4 \) into the second row equation: \[ 0\times x + 1\times y - 3\times 4 = -9 \Rightarrow y - 12 = -9 \Rightarrow y = 3 \].- Substitute \( z = 4 \) and \( y = 3 \) into the first row equation: \[ x + 3 + 4 = 6 \Rightarrow x + 7 = 6 \Rightarrow x = -1 \].

Step 4: Write the Solution as an Ordered Triple

The solution can be written based on the values we found. The ordered triple representing the solutions is:\[ (-1, 3, 4) \]

Key Concepts

System of Linear Equations

A system of linear equations consists of multiple linear equations, all of which have the same set of unknowns. In real-world applications, these must be solved to find shared values of these variables that satisfy all equations simultaneously.
Understanding this system is crucial because many real-world problems can be modeled through linear equations, from economic forecasts to engineering calculations.
In our exercise, the system involves three variables, \( x \), \( y \), and \( z \), and three equations, presenting a scenario where we're looking for values that keep all equations balanced:

• \( x + y + z = 6 \)
• \( 2x + 3y - z = 3 \)
• \( x + y + 2z = 10 \)

The goal is to find the values of \( x \), \( y \), and \( z \) that solve the system consistently.

Augmented Matrix

An augmented matrix is a powerful tool that represents systems of linear equations in matrix form. It aligns the coefficients of each variable in the equations and separates them by a line from the constants.
This form simplifies solving the system by allowing row operations, through which we can manipulate the rows to simplify the system and make the variables more accessible. Consider the augmented matrix for the given system:

• \( \begin{bmatrix} 1 & 1 & 1 & | & 6 \ 2 & 3 & -1 & | & 3 \ 1 & 1 & 2 & | & 10 \end{bmatrix} \)

This matrix includes:

• Columns for each variable, aligning the coefficients of \( x \), \( y \), and \( z \).
• The far-right column represents the constants from each equation.

Using this structured format, we're poised to apply Gaussian elimination to solve the system.

Row Operations

Row operations are procedures used to simplify matrices toward achieving row echelon form, crucial for solving systems via Gaussian elimination. They include:

• Swapping two rows.
• Multiplying a row by a non-zero constant.
• Adding or subtracting the multiple of one row from another.

These operations ensure equivalence of the linear equations while manipulating the rows during the elimination process.
In this exercise, we perform two key row operations to progress towards an upper triangular form:

• Subtracting the first row from the third to adjust its leading coefficient to zero: \( R_3 = R_3 - R_1 \).
• Eliminating the \( x \) coefficient in the second row through \( R_2 = R_2 - 2R_1 \).

Through careful application, these operations pave the way to easily identify values of the system's unknowns.

Backward Substitution

After achieving an upper triangular form via Gaussian elimination, backward substitution is the next step. It solves the system beginning from the last equation, where typically, only one variable is present. Doing so enables solving for that variable and substituting back to resolve other variables step by step.
With our exercise, backward substitution begins with
row 3, which directly gives us \( z = 4 \).

• Using this, substitute \( z = 4 \) into row 2: \( y - 3 \times 4 = -9 \), solving gives \( y = 3 \).
• Finally, substitute both \( y = 3 \) and \( z = 4 \) into row 1 to solve \( x + 3 + 4 = 6 \), providing \( x = -1 \).

This systematic approach uncovers the solution \((-1, 3, 4)\), completely solving our original system of equations.