Problem 18 - Algebra and Trigonometry

Question

15–24 The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$\left\\{\begin{aligned} x+y+z =4 \\\\-x+2 y+3 z =17 \\ 2 x-y \quad=-7 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = -2\), \(y = 3\), \(z = 3\).

Step 1: Set Up Augmented Matrix

First, express the system of equations as an augmented matrix: \[\begin{bmatrix}1 & 1 & 1 & | & 4 \-1 & 2 & 3 & | & 17 \2 & -1 & 0 & | & -7 \end{bmatrix}\]

Step 2: Use Row Operations to Achieve Upper Triangular Form

Perform row operations to create zeros under the first pivot element (1) in the first column.- Add Row 1 to Row 2 to get the new Row 2: - \(R_2 = R_2 + R_1\) - Resulting in \([0, 3, 4, |, 21]\)- Subtract 2 * Row 1 from Row 3: - \(R_3 = R_3 - 2R_1\) - Resulting in \([0, -3, -2, |, -15]\)Now our matrix looks like:\[\begin{bmatrix}1 & 1 & 1 & | & 4 \0 & 3 & 4 & | & 21 \0 & -3 & -2 & | & -15 \end{bmatrix}\]

Step 3: Eliminate Second Column's Second Row Terms

To eliminate the term in Row 3, Column 2, add Row 2 to Row 3:\(R_3 = R_3 + R_2\)This operation yields:\[\begin{bmatrix}1 & 1 & 1 & | & 4 \0 & 3 & 4 & | & 21 \0 & 0 & 2 & | & 6 \end{bmatrix}\]

Step 4: Solve for the Last Variable (z)

From the third row, \(2z = 6\). Solve for \(z\):\[z = \frac{6}{2} = 3\]

Step 5: Back Substitute to Solve for y

Using the second row, substitute \(z = 3\) into the equation:\[3y + 4(3) = 21\]\[3y + 12 = 21\]\[3y = 9\]\[y = \frac{9}{3} = 3\]

Step 6: Back Substitute to Solve for x

Finally, substitute \(y = 3\) and \(z = 3\) into the first row:\[x + 3 + 3 = 4\]\[x + 6 = 4\]\[x = 4 - 6\]\[x = -2\]

Step 7: State Final Answer

The solution to the system is \(x = -2\), \(y = 3\), and \(z = 3\).

Key Concepts

System of Linear Equations

A system of linear equations is a set of equations with multiple variables. Each equation represents a line or a plane in a multi-dimensional space. The solution to the system of linear equations is the set of values for the variables that satisfies all equations simultaneously.
In our example, we have the system: - \(x + y + z = 4\) - \(-x + 2y + 3z = 17\) - \(2x - y = -7\) These three equations form a system that can be visualized as three planes intersecting in 3D space. The point where all three planes meet is the solution to the system. In some cases, the system might have no solution, one solution (unique solution), or infinitely many solutions.

Augmented Matrix

To solve a system of linear equations, one effective method is to convert the set of equations into an augmented matrix. This matrix combines the coefficients of the variables and the constants from the equations.
An augmented matrix consists of rows and columns:

• Each row represents an equation.
• Each column represents a coefficient of a variable in the equations.
• An additional column is added for the constants on the right side of the equation, separated by a vertical bar \(|\).

For our example, the system of equations is expressed as the augmented matrix:\[\begin{bmatrix}1 & 1 & 1 & | & 4 \-1 & 2 & 3 & | & 17 \2 & -1 & 0 & | & -7\end{bmatrix}\]
This matrix simplifies the process of performing operations needed to find solutions by focusing on coefficients and constants without rewriting the variables.

Row Operations

Row operations are techniques used to manipulate the rows of an augmented matrix in order to solve a system of linear equations through Gaussian elimination. The goal is to achieve an upper triangular form, where zeros appear below the pivot elements, making it easier to solve for each variable. Row operations come in three types:

• Swap Rows: Exchange two rows with each other.
• Multiply a Row by a Scalar: Multiply all elements of a row by a non-zero constant.
• Add or Subtract Rows: Add or subtract a multiple of one row to another. This is used to eliminate coefficients systematically.

In our problem, row operations were performed to manipulate the augmented matrix into a simpler form:

• Add Row 1 to Row 2: \([R_2 = R_2 + R_1]\)
• Subtract 2 times Row 1 from Row 3: \([R_3 = R_3 - 2R_1]\)
• Add Row 2 to Row 3: \([R_3 = R_3 + R_2]\)

These operations gradually transform the matrix, allowing one to move towards isolating each variable and ultimately solving the system.