Problem 22 - College Algebra

Question

Solve the system by Gaussian elimination. \(6 x+2 y=-4\) \(3 x+4 y=-17\)

Step-by-Step Solution

Answer

The solution is \(x = \frac{15}{11}\) and \(y = -\frac{49}{11}\).

Step 1: Write the System in Matrix Form

First, transform the given system of equations into a matrix. We write the coefficients of the variables and the constants in matrix form: \[ \begin{bmatrix} 6 & 2 \ 3 & 4 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -4 \ -17 \end{bmatrix} \].

Step 2: Form the Augmented Matrix

Create the augmented matrix from the equations by writing the coefficients and constants together: \[ \begin{bmatrix} 6 & 2 & | & -4 \ 3 & 4 & | & -17 \end{bmatrix} \].

Step 3: Use Row Operations to Achieve Row Echelon Form

Our first goal is to create a leading '1' (pivot) in the first row, first column. We can do this by dividing the entire first row by 6:\[ \begin{bmatrix} 1 & \frac{1}{3} & | & -\frac{2}{3} \ 3 & 4 & | & -17 \end{bmatrix} \].Now eliminate the 3 below the pivot by replacing Row 2 with (Row 2 - 3 * Row 1):\[ \begin{bmatrix} 1 & \frac{1}{3} & | & -\frac{2}{3} \ 0 & \frac{11}{3} & | & -\frac{49}{3} \end{bmatrix} \].

Step 4: Simplify the Second Row

Multiply the entire second row by \( \frac{3}{11} \) to create a leading '1' in the second row, second column:\[ \begin{bmatrix} 1 & \frac{1}{3} & | & -\frac{2}{3} \ 0 & 1 & | & -\frac{49}{11} \end{bmatrix} \].

Step 5: Solve for the First Variable

Substitute the result from the second row back into the first to eliminate \( y \): Replace Row 1 with (Row 1 - \( \frac{1}{3} \)* Row 2):\[ \begin{bmatrix} 1 & 0 & | & \frac{15}{11} \ 0 & 1 & | & -\frac{49}{11} \end{bmatrix} \].From this matrix, we can read off the solutions directly: \( x = \frac{15}{11} \) and \( y = -\frac{49}{11} \).

Step 6: Finalize the Solution

Convert the results from the matrix back into equations:1. \( x = \frac{15}{11} \)2. \( y = -\frac{49}{11} \).These are the values of \( x \) and \( y \) that solve the system of equations.

Key Concepts

Matrix Form

When solving systems of linear equations, the first step is often to represent them in matrix form. This involves organizing the coefficients of the variables and the constants into a structured format like a grid. For example, the system:

• 6x + 2y = -4
• 3x + 4y = -17

can be expressed as the matrix equation:\[\begin{bmatrix} 6 & 2 \ 3 & 4 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -4 \ -17 \end{bmatrix}.\]This transformation to matrix form helps streamline further calculations and preparations for Gaussian Elimination. It serves the purpose of organizing the information needed to work on solving the equations systematically. By doing this first, it sets a clear foundation for the subsequent steps.

Augmented Matrix

Once you have the matrix form of the equations, the next step is to create an augmented matrix. An augmented matrix simply extends the original matrix by including the constants from the right-hand side of the equations. The format is as follows:\[\begin{bmatrix} 6 & 2 & | & -4 \ 3 & 4 & | & -17 \end{bmatrix}.\]Here, the vertical line separates the coefficients of the variables from the constants. The augmented matrix makes it easier to perform row operations as it lays out all necessary numbers to progress toward solving the system. Think of it as a complete summary of the problem in a compact form, ready for further manipulation.

Row Operations

Row operations are key moves in the method of Gaussian Elimination. These operations involve:

• Interchanging rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting free-of-error multiples of one row to another

In our example, to simplify the system, the first row is divided by 6:\[\begin{bmatrix} 1 & \frac{1}{3} & | & -\frac{2}{3} \ 3 & 4 & | & -17 \end{bmatrix}.\]Then, to remove elements below the pivot, the second row is updated by performing row operations such as replacing Row 2 with (Row 2 - 3 * Row 1). These steps manipulate the matrix strategically to streamline it into a form that's easier to interpret and solve.

Row Echelon Form

The ultimate goal with Gaussian Elimination is to reshape the matrix into Row Echelon Form (REF). This means that the matrix is transformed such that:

• All zeros are at the bottom of the matrix
• Each leading coefficient (or pivot) of a non-zero row is right of the leading coefficient of the row above it
• The entries below each pivot are zero

For example, after our performed row operations, the matrix becomes:\[\begin{bmatrix} 1 & 0 & | & \frac{15}{11} \ 0 & 1 & | & -\frac{49}{11} \end{bmatrix}.\]This is a simplified version of the system and is in a form where the values of the variables can be directly read off. In this case, the variables are already isolated, leading to the solution \( x = \frac{15}{11} \) and \( y = -\frac{49}{11} \). Thus, achieving REF is crucial as it simplifies solving the equations.