Problem 23 - College Algebra

Question

Solve each system by Gaussian elimination. $$ \begin{aligned} x+y+z &=14 \\ 2 y+3 z &=-14 \\ -16 y-24 z &=-112 \end{aligned} $$

Step-by-Step Solution

Answer

The system has infinitely many solutions depending on parameter \( t \).

Step 1: Write the Augmented Matrix

The given system of equations can be written in matrix form. The augmented matrix for the system is: \[ \begin{bmatrix} 1 & 1 & 1 & | & 14 \ 0 & 2 & 3 & | & -14 \ 0 & -16 & -24 & | & -112 \end{bmatrix} \].

Step 2: Perform Row Operations to Reach Row Echelon Form

To simplify the matrix, we focus on making zeros below the leading 1s. We start with the first column.Since the first element of the second row is already 0, we can focus on the third row. Perform the operation Row 3 = Row 3 + 8*Row 2 to eliminate the first element of Row 3:\[\begin{bmatrix} 1 & 1 & 1 & | & 14 \0 & 2 & 3 & | & -14 \0 & 0 & 0 & | & 0 \end{bmatrix}\].

Step 3: Interpret the Resulting Matrix

Examine the resulting matrix. The third row is all zeros, suggesting the system may have infinitely many solutions or a parameter dependency. The second row corresponds to an equation without any inconsistencies:\[2y + 3z = -14\].This implies that any solution involving y and z must satisfy this equation.

Step 4: Solve for the Remaining Variables

Now solve the system back-substituting from the equations given by the first and second rows:1. From row 1: \( x + y + z = 14 \).2. From row 2: \( 2y + 3z = -14 \).Assume \( z = t \) where \( t \) is a free variable. Substitute into the second equation:\[ 2y + 3t = -14 \]Solve for \( y \):\[ y = \frac{-14 - 3t}{2} \]Substitute \( y \) and \( z \) into the first equation:\[ x + \frac{-14 - 3t}{2} + t = 14 \]Simplify to find \( x \):\[ x = 14 - \frac{-14 - 3t}{2} - t \].

Key Concepts

Augmented Matrix

In Gaussian elimination, we use an augmented matrix to represent the system of linear equations. This is a compact way of organizing the coefficients and constants of a system, making it easier to apply row operations later.

The matrix consists of two parts:

• The coefficient matrix, which contains the coefficients of the variables.
• The augmented column, which includes the constants from each equation.

By combining these, we can perform operations without having to repeatedly write the variables. For example, the system\[\begin{aligned} x + y + z &=14 \2 y+3 z &=-14 \-16 y-24 z &=-112 \end{aligned}\]is expressed as the augmented matrix:\[\begin{bmatrix}1 & 1 & 1 & | & 14 \0 & 2 & 3 & | & -14 \0 & -16 & -24 & | & -112 \end{bmatrix}\]By laying out the equations in this form, you can visualize row operations more easily.

Row Operations

Row operations are fundamental to simplifying the augmented matrix into an easier-to-solve form, known as row echelon form. There are three types of row operations you can use:

• Interchanging two rows.
• Multiplying a row by a non-zero scalar.
• Adding or subtracting a multiple of one row to another.

These operations help us create zeros below the pivots—leading 1s—in a systematic way.

In the example system, we perform an operation on the third row to eliminate the coefficient beneath the first column's leading 1. This operation is:\[\text{Row 3} = \text{Row 3} + 8 \times \text{Row 2}\]These transformations simplify the system to:\[\begin{bmatrix}1 & 1 & 1 & | & 14 \0 & 2 & 3 & | & -14 \0 & 0 & 0 & | & 0 \end{bmatrix}\]The resulting matrix brings us closer to solving for specific variable values.

Row Echelon Form

The row echelon form is a simplified version of a matrix that is easier to understand and solve. It is achieved by applying row operations until each leading entry (non-zero) in a row is to the right of the leading entry in the row above, and all elements below leading entries are zeros.

This form allows you to systematically solve equations, often using back substitution. In our example, the transformed augmented matrix results in:\[\begin{bmatrix}1 & 1 & 1 & | & 14 \0 & 2 & 3 & | & -14 \0 & 0 & 0 & | & 0 \end{bmatrix}\]The zeros in the third row hint at special properties of our system, such as the possibility of infinite solutions. This form helps guide us in deciding the next steps in solving for variable values.

Infinite Solutions

When a system of equations leads to an augmented matrix with a row of all zeros, it may indicate infinitely many solutions. This situation arises due to dependency among the equations.

In such cases, one or more variables act as free variables, meaning they can take on any value that satisfies the remaining equations. For our example:

• From the second row, we have the equation: \( 2y + 3z = -14 \)
• From the first row, we get: \( x + y + z = 14 \)

We can choose one variable to be free—often denoted by a parameter, like \( z = t \)—allowing us to express other variables in terms of this parameter. This yields solutions in the form:\[\begin{aligned}y &= \frac{-14 - 3t}{2} \x &= 14 - \frac{-14 - 3t}{2} - t\end{aligned}\]This parameterized solution set reflects that any value of \( t \) producing \( x \), \( y \), and \( z \) still satisfies our original equations—demonstrating infinite solutions.