Problem 43 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system by Gaussian elimination. $$ \begin{array}{c} x+y+z=100 \\ x+2 z=125 \\ -y+2 z=25 \end{array} $$

Step-by-Step Solution

Answer

The solution is \(x = 91.67\), \(y = -\frac{25}{3}\), \(z = \frac{50}{3}\).

Step 1: Write the System as an Augmented Matrix

First, we'll convert the given system of equations into an augmented matrix. Each row in the matrix corresponds to the coefficients of one equation:\[\begin{bmatrix} 1 & 1 & 1 & | & 100 \ 1 & 0 & 2 & | & 125 \ 0 & -1 & 2 & | & 25 \end{bmatrix}\]

Step 2: Use Row Operations to Make the Matrix Upper Triangular

We'll use row operations to zero out the elements below the first leading 1 in the matrix. Start by subtracting the first row from the second row to eliminate the first element in the second row.- Replace Row 2: Row 2 - Row 1\[\begin{bmatrix} 1 & 1 & 1 & | & 100 \ 0 & -1 & 1 & | & 25 \ 0 & -1 & 2 & | & 25 \end{bmatrix}\]

Step 3: Eliminate Below the Second Pivot

Continue the process by eliminating the first element in Row 3 by using Row 2. Add Row 2 to Row 3:- Replace Row 3: Row 3 + Row 2\[\begin{bmatrix} 1 & 1 & 1 & | & 100 \ 0 & -1 & 1 & | & 25 \ 0 & 0 & 3 & | & 50 \end{bmatrix}\]

Step 4: Solve for z

Now that the matrix is in upper triangular form, solve for the last variable, \(z\), in the third row:- From Row 3: \(3z = 50\)- Thus, \(z = \frac{50}{3}\).

Step 5: Substitute z into Second Row to Solve for y

Use the value of \(z\) from Step 4 in the second row to solve for \(y\):- From Row 2: \(-y + \left(\frac{50}{3}\right) = 25\)- Simplify to find \(y\): \(-y = 25 - \frac{50}{3}\) Calculate: \(-y = \frac{75}{3} - \frac{50}{3} = \frac{25}{3}\)- Thus, \(y = -\frac{25}{3}\).

Step 6: Substitute y and z into First Row to Solve for x

Now substitute \(y = -\frac{25}{3}\) and \(z = \frac{50}{3}\) back into the first row to solve for \(x\):- From Row 1: \(x + \left(-\frac{25}{3}\right) + \left(\frac{50}{3}\right) = 100\)- Simplify to find \(x\): \(x + \frac{25}{3} = 100\) \(x = 100 - \frac{25}{3}\) \(x = 100 - 8.33 = 91.67\).

Key Concepts

System of Equations

A system of equations consists of multiple equations aimed at finding common solutions for variable values. In this case, we encounter a 3-variable system with equations:

• \(x + y + z = 100\)
• \(x + 2z = 125\)
• \(-y + 2z = 25\)

These equations describe a relationship between the variables \(x\), \(y\), and \(z\). The objective is to find a set of values for these variables that satisfy all three equations simultaneously. Solving the system provides insights into how these variables interact and depend on each other in this specific context. Systems of equations can be solved using various methods, such as substitution, elimination, or matrix techniques like Gaussian elimination.

Augmented Matrix

The augmented matrix is a crucial step in solving a system of equations using linear algebra techniques. It captures the structure of the equations in a compact and efficient manner. By transforming the given system:

• \(x + y + z = 100\)
• \(x + 2z = 125\)
• \(-y + 2z = 25\)

into the matrix form: \[\begin{bmatrix} 1 & 1 & 1 & | & 100 \ 1 & 0 & 2 & | & 125 \ 0 & -1 & 2 & | & 25 \end{bmatrix}\]The coefficients of the variables are aligned in columns, and the constants are separated by a '|' to keep track of the operations. This format makes it easier to perform systematic operations on the entire system rather than dealing with individual equations. The ultimate goal here is to transform the matrix into a form which simplifies finding the solutions for the variables.

Row Operations

Row operations are the fundamental tools we use to manipulate the augmented matrix. There are three types of row operations:

• Swapping two rows
• Multiplying a row by a nonzero constant
• Adding or subtracting rows

These operations allow us to change the matrix’s appearance to solve for the variables, without altering the underlying relationships expressed by the system. Starting with the system's augmented matrix, a series of row operations are applied to achieve the easier-to-solve upper triangular form. This includes:

• First, subtracting the first row from the second row, transforming the matrix to:
\[\begin{bmatrix} 1 & 1 & 1 & | & 100 \ 0 & -1 & 1 & | & 25 \ 0 & -1 & 2 & | & 25 \end{bmatrix}\]
• Adding the modified second row to the third row, which further simplifies computation.

Through this method, we control the elements of the matrix to reveal the solutions in later steps.

Upper Triangular Matrix

An upper triangular matrix is a simplified form where all the elements below the leading diagonal (from top left to bottom right) are zeros. For this system, restructuring the augmented matrix into an upper triangular matrix involves:Transforming \[\begin{bmatrix} 1 & 1 & 1 & | & 100 \ 0 & -1 & 1 & | & 25 \ 0 & 0 & 3 & | & 50 \end{bmatrix}\]This format is advantageous because it isolates variables gradually, allowing for straightforward back-substitution to solve for one variable at a time. We begin from the last equation - often containing just one variable - and substitute back incrementally to find the values of all other variables. This systematic reduction reveals the specific values for \(x\), \(y\), and \(z\) efficiently. The matrix manipulation ensures a step-by-step approach that reduces complexity and clarifies the solution process.