Problem 22 - Precalculus: Mathematical for Calculus

Question

The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$\left\\{\begin{array}{lr} 2 x_{1}+x_{2} & =7 \\ 2 x_{1}-x_{2}+x_{3} & =6 \\ 3 x_{1}-2 x_{2}+4 x_{3} & =11 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \(x_1 = 3\), \(x_2 = -\frac{2}{5}\), \(x_3 = \frac{6}{5}\).

Step 1: Write the Augmented Matrix

First, we need to write the augmented matrix that represents the system of equations. The given equations are:\[\begin{align*}2x_1 + x_2 &= 7, \2x_1 - x_2 + x_3 &= 6, \3x_1 - 2x_2 + 4x_3 &= 11.\end{align*}\]The augmented matrix is:\[\begin{bmatrix}2 & 1 & 0 & | & 7 \2 & -1 & 1 & | & 6 \3 & -2 & 4 & | & 11\end{bmatrix}\]

Step 2: Make the Leading Coefficient of the First Row 1

We need the leading coefficient (first element) of the first row to be 1. In this matrix, the first element is already 2, so divide the entire first row by 2:\[\begin{bmatrix}1 & \frac{1}{2} & 0 & | & \frac{7}{2} \2 & -1 & 1 & | & 6 \3 & -2 & 4 & | & 11\end{bmatrix}\]

Step 3: Eliminate the First Column Below the First Row

To eliminate the first column elements below the pivot (which is 1), subtract 2 times the first row from the second row, and subtract 3 times the first row from the third row:\[\begin{bmatrix}1 & \frac{1}{2} & 0 & | & \frac{7}{2} \0 & -2 & 1 & | & \frac{1}{2} \0 & -\frac{7}{2} & 4 & | & \frac{5}{2}\end{bmatrix}\]

Step 4: Make the Leading Coefficient of the Second Row 1

To make the leading coefficient (the first nonzero element of the second row) 1, divide the second row by \(-2\):\[\begin{bmatrix}1 & \frac{1}{2} & 0 & | & \frac{7}{2} \0 & 1 & -\frac{1}{2} & | & -\frac{1}{4} \0 & -\frac{7}{2} & 4 & | & \frac{5}{2}\end{bmatrix}\]

Step 5: Eliminate the Second Column in the Third Row

Subtract \(\frac{7}{2}\) times the second row from the third row to eliminate the second column of the third row:\[\begin{bmatrix}1 & \frac{1}{2} & 0 & | & \frac{7}{2} \0 & 1 & -\frac{1}{2} & | & -\frac{1}{4} \0 & 0 & \frac{5}{2} & | & 3\end{bmatrix}\]

Step 6: Make the Leading Coefficient of the Third Row 1

Divide the third row by \(\frac{5}{2}\) to make the leading coefficient of the third row 1:\[\begin{bmatrix}1 & \frac{1}{2} & 0 & | & \frac{7}{2} \0 & 1 & -\frac{1}{2} & | & -\frac{1}{4} \0 & 0 & 1 & | & \frac{6}{5}\end{bmatrix}\]

Step 7: Back Substitution to Find Unique Solution

Now perform back substitution:1. Substitute the value from the third row up to eliminate the third column from the second row:\[\begin{bmatrix}1 & \frac{1}{2} & 0 & | & \frac{7}{2} \0 & 1 & 0 & | & -\frac{2}{5} \0 & 0 & 1 & | & \frac{6}{5}\end{bmatrix}\]2. Replace the third column's influence on the first row:\[\begin{bmatrix}1 & 0 & 0 & | & 3 \0 & 1 & 0 & | & -\frac{2}{5} \0 & 0 & 1 & | & \frac{6}{5}\end{bmatrix}\]Thus, the solution is \(x_1 = 3\), \(x_2 = -\frac{2}{5}\), \(x_3 = \frac{6}{5}\).

Step 8: Write the Solution

The solution to the system of equations is:\[(x_1, x_2, x_3) = \left(3, -\frac{2}{5}, \frac{6}{5}\right)\]

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more linear equations that are simultaneously true. Each equation represents a linear relationship between some unknown variables. The goal is to find values for these variables that make all the equations true at the same time. In our problem, we have three equations with unknowns: \(x_1, x_2,\) and \(x_3\). Each equation can be interpreted as a line or a plane in a geometric space. Solving the system means finding the intersection of these lines or planes.
Our system includes:

• \(2x_1 + x_2 = 7\)
• \(2x_1 - x_2 + x_3 = 6\)
• \(3x_1 - 2x_2 + 4x_3 = 11\)

Depending on the system, the solution can be:

• Unique solution: One set of values that satisfies all equations (as in this exercise).
• No solution: The equations contradict each other.
• Infinite solutions: Many sets of values satisfy all equations. This happens when equations describe the same line or plane.

Understanding these solutions helps in determining the system's behavior and predicting outcomes.

Augmented Matrix

The augmented matrix is a simple way to organize and solve a system of linear equations. It combines the coefficients of variables with their corresponding constants into a single matrix. This method is beneficial for executing operations like those in Gaussian elimination without dealing with equations directly. For our equations, the augmented matrix is:
\[\begin{bmatrix}2 & 1 & 0 & | & 7 \ 2 & -1 & 1 & | & 6 \ 3 & -2 & 4 & | & 11\end{bmatrix}\]
The vertical line separates the coefficients from the constants. Each row corresponds to one of the linear equations:

• First row represents \(2x_1 + x_2 = 7\)
• Second row represents \(2x_1 - x_2 + x_3 = 6\)
• Third row represents \(3x_1 - 2x_2 + 4x_3 = 11\)

Using this matrix, the system can be solved through row operations, making the process systematic and less error-prone compared to solving individual equations.

Back Substitution

Back substitution is an essential step in the Gaussian elimination method. Once a system of equations is simplified to an upper triangular form, where all coefficients below the main diagonal are zeros, back substitution is employed to find the solution.
In the context of our example, we start once the matrix has been transformed to:
\[\begin{bmatrix}1 & 0 & 0 & | & 3 \ 0 & 1 & 0 & | & -\frac{2}{5} \ 0 & 0 & 1 & | & \frac{6}{5}\end{bmatrix}\]
At this stage, each row of the matrix corresponds to a statement about each unknown:

• \(x_3 = \frac{6}{5}\)
• \(x_2 = -\frac{2}{5}\)
• \(x_1 = 3\)

The purpose of back substitution is simple: Solve for the variables starting from the bottom row and work upwards, substituting known values into the equations above. It untangles the variables step-by-step, leading to a final solution set, which in our case is \((x_1, x_2, x_3) = (3, -\frac{2}{5}, \frac{6}{5})\). This step locks in the unique values for each variable, confirming the solution's accuracy.