Problem 54 - Precalculus

Question

Solve the system of linear equations using Gaussian elimination with back- substitution. $$\begin{aligned} &3 x-y=-1\\\ &2 y+6 x=2 \end{aligned}$$

Step-by-Step Solution

Answer

The solution of the system is \((0, 1)\).

Step 1: Write the system in standard form

We start by writing the system of linear equations in standard form. Rearrange both equations to align the terms by variable:\[ 3x - y = -1 \]\[ 6x + 2y = 2 \]

Step 2: Form the augmented matrix

Convert the system of equations into an augmented matrix:\[\begin{bmatrix}3 & -1 & | & -1 \6 & 2 & | & 2\end{bmatrix}\]

Step 3: Perform row operations to reach row echelon form

Our goal is to create zeros below the first pivot (3 in the first row, first column). To eliminate the 6 beneath it, perform the operation: Row 2 = Row 2 - 2*Row 1:\[\begin{bmatrix}3 & -1 & | & -1 \0 & 4 & | & 4\end{bmatrix}\]

Step 4: Solve for the second variable

With the matrix in row echelon form, solve the second equation in the transformed matrix to find the value of \( y \):\[4y = 4\]Dividing both sides by 4 gives:\[y = 1\]

Step 5: Back-substitute to solve for the first variable

Substitute \( y = 1 \) into the first equation to solve for \( x \):\[3x - 1 = -1\]Adding 1 to both sides gives:\[3x = 0\]Divide by 3:\[x = 0\]

Step 6: Write the solution as an ordered pair

The solution to the system of equations is given by the values of \( x \) and \( y \):\[(x, y) = (0, 1)\]

Key Concepts

Understanding Linear Equations

Linear equations form the cornerstone of countless mathematical problems. They involve expressions where the variables are only raised to the power of one. This simplicity makes them linear, as opposed to quadratic or higher-degree polynomial equations, which are non-linear in nature.
Each equation generally takes the form of \( ax + by + cz + \ldots = d \), where \( a, b, c, \) and \( d \) are constants, and \( x, y, z \) are variables.
In our example, we start with two equations:

• \( 3x - y = -1 \)
• \( 6x + 2y = 2 \)

These represent lines in a two-dimensional space when plotted on a graph. The solution (if it exists) is the point where these lines intersect. The goal is to find this intersection point, which in this context is represented by the values of \( x \) and \( y \).

Exploring Augmented Matrix

An augmented matrix is a helpful tool for solving systems of linear equations, particularly when using methods like Gaussian elimination.
It combines the coefficients of the variables and the constant terms of the equations into a single matrix form.

• The coefficient matrix captures all the coefficients of the variables from the equations.
• The constant column includes the standalone numbers from each equation.

For our system of equations, the augmented matrix is:\[\begin{bmatrix}3 & -1 & | & -1 \6 & 2 & | & 2\end{bmatrix}\]The vertical bar distinguishes the coefficients on the left from the constants on the right side of the equations, maintaining the structure we saw in the original system. By structuring the system as a matrix, we can efficiently perform row operations needed for Gaussian elimination.

Achieving Row Echelon Form

The row echelon form of a matrix streamlines the process of solving linear equations. It aims to simplify the matrix to a state where back-substitution becomes possible.
To reach this form, we perform row operations on the augmented matrix:

• Use row interchange, scalar multiplication, and row addition.
• Focus on creating leading 1s in each row and zeros below them.

In our example, the first step was to eliminate the leading coefficient beneath 3 in the first column.
This was done by the operation: Row 2 = Row 2 - 2*Row 1. This simplified our matrix to:\[\begin{bmatrix}3 & -1 & | & -1 \0 & 4 & | & 4\end{bmatrix}\]Now, our matrix is in row echelon form, making it easier to solve for the variables through back-substitution.

Executing Back-Substitution

Back-substitution is the final step in solving linear systems through Gaussian elimination. It is used once we have a matrix in row echelon form, allowing us to solve for variables from the bottom up.
Begin with the last row, which should ideally depict just one variable, and solve for that variable.

• In our matrix: \(4y = 4\) translates to \(y = 1\).
• Substitute \(y = 1\) back into the equation derived from the first row.- This equation: \(3x - 1 = -1\) simplifies to \(3x = 0\), giving \(x = 0\).

This method ensures that each variable is determined successively, using information from the rows below it. Consequently, the solution to the system is the ordered pair \((x, y) = (0, 1)\), where both equations harmonize. Thus, the solution represents the precise point of intersection of the lines that the original equations describe.