Problem 23 - College Algebra with Corequisite Support

Question

For the following exercises, solve each system by Gaussian elimination. $$ \begin{array}{c} x+y+z=14 \\ 2 y+3 z=-14 \\ -16 y-24 z=-112 \end{array} $$

Step-by-Step Solution

Answer

System has infinite solutions: \( x = \frac{42+t}{2}, \ y = \frac{-14-3t}{2}, \ z = t \).

Step 1: Write the augmented matrix

First, write the given system of equations in the form of an augmented matrix. This will help us systematically apply Gaussian elimination. The augmented matrix is:\[\begin{bmatrix}1 & 1 & 1 & | & 14 \0 & 2 & 3 & | & -14 \0 & -16 & -24 & | & -112 \\end{bmatrix}\]

Step 2: Make the second row entry zero in the first column

Since the first column already has zeros below the first row, we don't need to modify it. The matrix remains the same:\[\begin{bmatrix}1 & 1 & 1 & | & 14 \0 & 2 & 3 & | & -14 \0 & -16 & -24 & | & -112 \\end{bmatrix}\]

Step 3: Eliminate the third row using the second row

To eliminate the first entry of the third row, we will perform: Row 3 = Row 3 + 8 * Row 2. The matrix becomes:\[\begin{bmatrix}1 & 1 & 1 & | & 14 \0 & 2 & 3 & | & -14 \0 & 0 & 0 & | & 0 \\end{bmatrix}\]

Step 4: Simplify and interpret the resulting system

We now interpret the resulting matrix as a system of equations:1. \( x + y + z = 14 \)2. \( 2y + 3z = -14 \)3. The third row zero indicates a dependent equation, implying infinite solutions. Describing one variable in terms of the other(s):- Solve equation 2 for one variable, say \( y \): \[ y = \frac{-14 - 3z}{2} \].4. Substitute into equation 1 for a general solution involving \( z \): \[ x + \left( \frac{-14 - 3z}{2} \right) + z = 14 \].

Step 5: Solve for the remaining variable

Rearrange and simplify equation 1 (\( x \) in terms of \( z \)):- Replace \( y \): \[ x - \frac{14 + 3z}{2} + z = 14 \].- Multiply through by 2 to clear the fraction: \[ 2x - (14 + 3z) + 2z = 28 \].- Simplify: \[ 2x -14 - z = 28 \]. \[ 2x = 42 + z \], or \[ x = \frac{42 + z}{2} \].Let \( z = t \), a parameter:\[ y = \frac{-14 - 3t}{2} \], \[ x = \frac{42 + t}{2} \].

Step 6: Express the general solution

The system has infinite solutions, parameterized by \( z = t \). The solution is:- \( x = \frac{42 + t}{2} \),- \( y = \frac{-14 - 3t}{2} \),- \( z = t \),where \( t \) is any real number.

Key Concepts

Augmented Matrix

An augmented matrix is a powerful tool to solve systems of linear equations using methods like Gaussian elimination. It combines the coefficients of variables and the constants from the right side of the equations into a single matrix, making calculations more straightforward. The matrix includes:

• The coefficients of all variables in the system, arranged in rows according to each equation.
• A vertical line dividing the coefficients from the constants, which represents the equal sign in the system of equations.

This setup allows us to apply matrix operations easily, helping simplify the system and find solutions. For instance, the given system of equations is written as:\[\begin{bmatrix}1 & 1 & 1 & | & 14 \0 & 2 & 3 & | & -14 \0 & -16 & -24 & | & -112 \\end{bmatrix}\]Using this matrix, we can quickly identify the necessary row operations to simplify the system, ultimately making the process of solving it more efficient.

Infinite Solutions

A system of equations may have infinite solutions, particularly when the equations are dependent or equivalent. This indicates that there isn't just one unique solution but rather a whole set or line of solutions.In our original problem, the Gaussian elimination resulted in a row of all zeros (\[0x + 0y + 0z = 0\]). This scenario reveals redundancy in the equations, pointing to a dependency that allows for multiple solutions.Here's why:

• The row of zeros shows that one equation does not add new information to the system.
• Other equations can be rewritten to express one or more variables in terms of free variables.

In this context, 'free variables' are those we can choose values for, which will then determine the other variables. This is why, in our solution, we express the variables using a parameter \(z = t\), which leads to multiple valid solutions depending on \(t\) (a real number).

System of Equations

A system of equations consists of two or more equations involving the same set of variables. Solving them means finding values for each variable that makes all equations true simultaneously. Depending on the system, the solution can be:

• Unique: A single set of values that satisfies all the equations.
• Infinite solutions: Many sets of values, often due to dependency between equations.
• No solution: When the equations are contradictory.

In our specific exercise, we started with a system of equations involving three variables \(x, y, \) and \(z\):\[\begin{align*}& x + y + z = 14 \& 2y + 3z = -14 \& -16y -24z = -112\end{align*}\]Using Gaussian elimination, we found that the equations were dependent, leading to infinite solutions. This method systematically reduces the complexity of the system by row operations to reach a conclusion, ensuring that we understand the relationship among the variables clearly. In the end, we express solutions in terms of one parameter, efficiently capturing all possible solutions.