Problem 19 - College Algebra

Question

For the following exercises, solve the system by Gaussian elimination. $$ \left[\begin{array}{rr|r}{-1} & {2} & {-3} \\ {4} & {-5} & {6}\end{array}\right] $$

Step-by-Step Solution

Answer

The solution is \( x = -1 \), \( y = -2 \).

Step 1: Identify the Augmented Matrix

The given exercise is already represented as an augmented matrix: \[\begin{bmatrix} -1 & 2 & | & -3 \ 4 & -5 & | & 6 \end{bmatrix}\] This signifies a system of linear equations that needs to be solved using Gaussian elimination.

Step 2: Convert First Row Element to Leading 1

To start Gaussian elimination, the first element of the first row should be 1. Multiply the entire first row by -1 to achieve this:\[\text{Row 1 becomes } 1\cdot(-1) = 1\\begin{bmatrix} 1 & -2 & | & 3 \ 4 & -5 & | & 6 \end{bmatrix}\]

Step 3: Eliminate First Column Elements Below Leading Coefficient

To cancel out the 4 beneath the leading 1 in column 1, modify Row 2 by subtracting 4 times Row 1 from it:\[\text{New Row 2} = ext{Row 2} - 4\cdot\text{Row 1}\]\[= \left[ 4, -5, | ,6 \right] - 4 \left[ 1, -2, | ,3 \right]\]\[= \left[ 4-4\cdot1, -5-4\cdot(-2), 6-4\cdot3 \right]\]\[= \begin{bmatrix} 1 & -2 & | & 3 \ 0 & 3 & | & -6 \end{bmatrix}\]

Step 4: Convert Second Row's Leading Coefficient to 1

In the second row equation, change the leading coefficient to 1 by dividing the entire second row by 3:\[\begin{bmatrix} 1 & -2 & | & 3 \ 0 & 1 & | & -2 \end{bmatrix}\]

Step 5: Back-substitute to Clear Upper Rows

Use the new leading coefficient in Row 2 to ensure 0s above it in Row 1 by adjusting Row 1:\[\text{Row 1} = ext{Row 1} + 2\cdot\text{Row 2}\]\[= \left[ 1, -2, | ,3 \right] + 2\cdot\left[ 0, 1, | , -2 \right]\]\[= \left[ 1, 0, | ,3 - 4 \right]\]\[= \left[1, 0, |, -1 \right]\]\[\text{The matrix now becomes } \begin{bmatrix} 1 & 0 & | & -1 \ 0 & 1 & | & -2 \end{bmatrix}\]

Step 6: Identify Solution from Row-Echelon Form

The matrix is now in row-echelon form. We can identify the variables as follows: - Row 1 suggests \( x = -1 \)- Row 2 suggests \( y = -2 \)Thus, the solution to the system of equations is \( x = -1 \) and \( y = -2 \).

Key Concepts

Augmented Matrix

An augmented matrix is a useful concept when dealing with systems of linear equations. Basically, it is a matrix that combines both the coefficient matrix and the constant terms from the equations into one compact form. This consolidated structure assists in simplifying computations, especially when applying methods such as Gaussian elimination.

For instance, consider the system represented in the problem:

• Equation 1: \(-x + 2y = -3\)
• Equation 2: \(4x - 5y = 6\)

To create an augmented matrix, organize the coefficients of the variables and the constants into rows:\[ \begin{bmatrix} -1 & 2 & | & -3 \ 4 & -5 & | & 6 \end{bmatrix} \]Each row corresponds to an equation, with the vertical line demarcating where coefficients end and constants begin. This approach enables seamless application of matrix operations to solve the equations.

Row Echelon Form

Row Echelon Form (REF) is an arrangement of an augmented matrix that significantly aids in solving systems of linear equations. The main idea is to have each row start with zeros followed by 1s (known as leading coefficients), moving from left to right, while maintaining zeros below each leading 1.

The Gaussian elimination process involves transforming the matrix to achieve the REF:

• The first step is to ensure the top left value becomes a 1. This is achieved by either swapping rows or multiplying a row by a constant.
• Next, create zeros below this leading 1 to cancel out coefficients in the same column. This often involves adding or subtracting multiples of the row.

In our example, we adjusted the equations via elimination and row manipulation to obtain the matrix:\[ \begin{bmatrix} 1 & 0 & | & -1 \ 0 & 1 & | & -2 \end{bmatrix} \]The matrix clearly shows the solutions; the top row detailing \(x = -1\), and the bottom row \(y = -2\), leading to straightforward back-substitution when needed.

System of Linear Equations

A System of Linear Equations consists of two or more linear equations that share two or more unknowns. The aim is to find the set of values for these unknowns that simultaneously satisfy each of the equations in the system.

The given system:

• Equation 1: \(-x + 2y = -3\)
• Equation 2: \(4x - 5y = 6\)

represents such a system where we need to find values for \(x\) and \(y\). By converting the system into an augmented matrix, we can apply Gaussian elimination to systematically reduce the equations to a simpler form, enabling us to readily deduce the solution.

By performing row operations, the system simplifies the process to determine the values of \(x\) and \(y\). This structured approach ensures precision, reducing the risk of calculation errors while providing a clear path to the solution. Here, solving the system yields \(x = -1\) and \(y = -2\), satisfying both equations. This method is key for students learning linear algebra as it builds foundational skills for more complex mathematical problem-solving.