Problem 15 - College Algebra

Question

Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$\begin{aligned}&2 x+y-z=2\\\&3 x+3 y-2 z=3\end{aligned}$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x=z-1, y=z/3+1/3\), where z can be any real number.

Step 1: Write the System as an Augmented Matrix

First the system of equations needs to be written as an augmented matrix. The augmented matrix for the given system is \[ \left[ \begin{array}{ccc|c} 2 & 1 & -1 & 2 \ 3 & 3 & -2 & 3 \end{array} \right] \].

Step 2: Convert to Row-Echelon Form

The next step is to convert the augmented matrix to row-echelon form. This will help in finding the solution easier. Subtract \(1.5\) times the first row from the second row to eliminate the \(x\) term from the second equation. The resulting matrix is \[ \left[ \begin{array}{ccc|c} 2 & 1 & -1 & 2 \ 0 & 1.5 & -0.5 & 0.5 \end{array} \right] \].

Step 3: Simplify the Matrix

Divide the second row by \(1.5\) to simplify the final matrix to \[ \left[ \begin{array}{ccc|c} 2 & 1 & -1 & 2 \ 0 & 1 & -1/3 & 1/3 \end{array} \right] \]. Use this reduced matrix to write the system of equations and solve for the variables.

Step 4: Solve for the variables

From the second row of the reduced matrix, we have \(y - z/3 = 1/3\). Simplifying it gives \(y=z/3+1/3\). Substituting \(y\) into the first equation, we get \( 2x + z + z = 2 \), which simplifies to \(x=z-1\). Hence the solution is \(x=z-1, y=z/3+1/3\), where z can be any real number.

Key Concepts

Augmented Matrix

An augmented matrix is an essential tool in solving a system of equations using Gaussian elimination. It combines the coefficients of the variables and the constants in a single matrix representation. This matrix makes it easy to apply row operations and visualize the steps needed to reach the solution.

For the given system of equations: \(2x + y - z = 2\) and \(3x + 3y - 2z = 3\), the augmented matrix is:

• Write down the coefficients of each variable in respective columns.
• Include constants as a separate column.

Thus, we have the matrix \( \left[ \begin{array}{ccc|c} 2 & 1 & -1 & 2 \ 3 & 3 & -2 & 3 \end{array} \right] \). Here, each row represents one equation from the system.

Row-Echelon Form

To efficiently solve a system of equations, the augmented matrix is converted into row-echelon form. This form simplifies the system and helps identify solutions.

• Each leading entry (first non-zero number from the left) in a row is to the right of the leading entry in the row above.
• All entries below a leading entry are zero.

In our example, we first perform a row operation: subtracting \(1.5\) times the first row from the second. This aims to eliminate the \(x\) term in the second equation. The resulting matrix is \( \left[ \begin{array}{ccc|c} 2 & 1 & -1 & 2 \ 0 & 1.5 & -0.5 & 0.5 \end{array} \right] \). Continuing with simplification, we divide the second row by \(1.5\) to further clear fractions, resulting in \( \left[ \begin{array}{ccc|c} 2 & 1 & -1 & 2 \ 0 & 1 & -1/3 & 1/3 \end{array} \right] \), which is now in row-echelon form.

System of Equations

Systems of equations consist of two or more equations that are solved together because they share variables.

• Each equation represents a condition that must be met simultaneously with the others.
• The goal is to find a set of values for the variables that satisfies all equations in the system.

In our starting equations: \(2x + y - z = 2\) and \(3x + 3y - 2z = 3\), these linear equations share the variables \(x\), \(y\), and \(z\). Through Gaussian elimination, the relationships between these variables are clarified, allowing us to find one or more solutions.

Solution of Equations

After converting the matrix to row-echelon form, the next step is solving for each variable. The simplified matrix

• Shows how each variable relates to others.
• Allows back-substitution to find specific values.

From the second row, \(y - \frac{z}{3} = \frac{1}{3}\), simplifying gives us \(y = \frac{z}{3} + \frac{1}{3}\). By substituting this \(y\) into the first equation, \(2x + y - z = 2\), we find \(x = z - 1\). The solution to the system is parametric, meaning \(x = z - 1\), \(y = \frac{z}{3} + \frac{1}{3}\), where \(z\) can be any real number. This represents an infinite set of solutions along a line in three-dimensional space.