Problem 72 - Precalculus

Question

Solve the system of linear equations using Gauss-Jordan elimination. $$\begin{aligned} &5 x-4 y=31\\\ &3 x+7 y=-19 \end{aligned}$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{109}{47}\) and \(y = -\frac{164}{47}\).

Step 1: Set up the augmented matrix

Begin by writing the system of equations as an augmented matrix. The first equation is \(5x - 4y = 31\) and the second equation is \(3x + 7y = -19\). The corresponding augmented matrix is: \[\begin{bmatrix}5 & -4 & | & 31 \3 & 7 & | & -19\end{bmatrix}\]

Step 2: Make the leading coefficient of the first row 1

To simplify the first equation, divide the first row by 5, making the leading coefficient 1.\[\begin{bmatrix}1 & -\frac{4}{5} & | & \frac{31}{5} \3 & 7 & | & -19\end{bmatrix}\]

Step 3: Eliminate the first column in the second row

Subtract 3 times the first row from the second row to eliminate the 3 in the second row, first column:\[\begin{bmatrix}1 & -\frac{4}{5} & | & \frac{31}{5} \0 & \frac{47}{5} & | & -\frac{164}{5}\end{bmatrix}\]

Step 4: Make the leading coefficient of the second row 1

Divide the entire second row by \(\frac{47}{5}\) to get a leading coefficient of 1:\[\begin{bmatrix}1 & -\frac{4}{5} & | & \frac{31}{5} \0 & 1 & | & -\frac{164}{47}\end{bmatrix}\]

Step 5: Eliminate the second column in the first row

Add \(\frac{4}{5}\) times the second row to the first row to eliminate \(-\frac{4}{5}\) in the first row, second column:\[\begin{bmatrix}1 & 0 & | & \frac{109}{47} \0 & 1 & | & -\frac{164}{47}\end{bmatrix}\]

Step 6: Write the solutions from the matrix

From the final row-reduced matrix, we interpret the values of \(x\) and \(y\) directly as: \(x = \frac{109}{47}\) and \(y = -\frac{164}{47}\).

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more equations with the same set of variables. In this case, we are dealing with two equations involving two variables, namely \(x\) and \(y\). The goal is to find values for these variables that satisfy both equations simultaneously. This often involves considering whether the lines represented by these equations intersect at a particular point.For example, in the given exercise:

• The first equation is \(5x - 4y = 31\).
• The second equation is \(3x + 7y = -19\).

By solving these equations together, we aim to find the particular values of \(x\) and \(y\) that make both equations true. This approach can be visualized graphically as finding the intersection point of the two lines plotted on a coordinate plane.

Augmented Matrix

To solve a system of linear equations using methods such as Gauss-Jordan elimination, we start by expressing the system in the form of an augmented matrix. An augmented matrix is a compact way to represent a system of linear equations as it combines the coefficients and constants of the equations into a unified matrix form.In our example, the system given by the equations

• \(5x - 4y = 31\)
• \(3x + 7y = -19\)

is turned into the following augmented matrix:\[\begin{bmatrix}5 & -4 & | & 31 \3 & 7 & | & -19\end{bmatrix}\]Each row of the matrix represents an equation, and each column (except the last) corresponds to the coefficients of a variable. The column to the right of the vertical bar represents the constants on the other side of the equations.

Row Operations

Row operations are key to manipulating augmented matrices and solving systems of equations. These operations include row swapping, scaling a row by a non-zero scalar, and adding or subtracting a multiple of one row to another.

Types of Row Operations

• **Swap Rows**: This operation exchanges one row with another to help position pivot elements where needed.
• **Scale Rows**: Multiply all entries of a row by a non-zero scalar to adjust the coefficients for easier elimination.
• **Add/Subtract Rows**: Replace a row by adding a multiple of another row to it, facilitating the elimination of coefficients.

In our Gauss-Jordan elimination process:

• We first scaled the first row to make the leading coefficient 1.
• Then, we eliminated other coefficients in the first column by adding/subtracting rows.
• Finally, we made adjustments to isolate coefficients and form a diagonal of 1s across the leading positions of each row.

These operations help convert the matrix into a simpler form, ultimately leading to finding solutions directly.

Solutions of Equations

The final step of the Gauss-Jordan elimination is reaching a form where the solutions can be read directly from the matrix. This matrix is called the reduced row echelon form (RREF), where each leading entry of a row is 1, and it is the only non-zero entry in its column.

Extracting Solutions

From the transformed matrix:\[\begin{bmatrix}1 & 0 & | & \frac{109}{47} \0 & 1 & | & -\frac{164}{47}\end{bmatrix}\]We can interpret the results as:

• \(x = \frac{109}{47}\)
• \(y = -\frac{164}{47}\)

These values provide the solutions to the original system of equations, indicating that the lines intersect at precisely these coordinates \((x, y)\). This means these are the values of \(x\) and \(y\) that satisfy both equations in the system simultaneously.