Problem 18 - College Algebra

Question

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

Step-by-Step Solution

Answer

The solution is \((-1, 2)\).

Step 1: Setup the Augmented Matrix

We begin with the given augmented matrix: \[ \begin{bmatrix} 1 & 2 & | & 3 \ 4 & 5 & | & 6 \end{bmatrix} \]. This represents the system of equations: \( x + 2y = 3 \) and \( 4x + 5y = 6 \).

Step 2: Eliminate x from the Second Equation

Use the first row to eliminate \( x \) from the second equation. Multiply the first row by 4 and subtract it from the second row. First row multiplied by 4: \( 4 \times (1, 2, |, 3) \) becomes \( (4, 8, |, 12) \).Subtract from the second row: \[ 4x + 5y = 6 \] becomes \[ (4-4)x + (5-8)y = (6-12) \] or \(-3y = -6\).

Step 3: Solve for y

From the simplified second equation \( -3y = -6 \), solve for \( y \) by dividing both sides by -3: \[ y = 2 \].

Step 4: Substitute y back into First Equation to Find x

Substitute \( y = 2 \) into the first equation \( x + 2y = 3 \): \[ x + 2(2) = 3 \]. Simplify this to obtain \( x = 3 - 4 = -1 \).

Step 5: Write the Solution

The solution to the system is \( x = -1 \) and \( y = 2 \). Therefore, the solution as an ordered pair is \((-1, 2)\).

Key Concepts

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. Instead of writing out each equation separately, we place all of their coefficients and the constants into a matrix. This makes it easier to apply techniques like Gaussian elimination.
In our example, the augmented matrix is given by: \[ \begin{bmatrix} 1 & 2 & | & 3 \ 4 & 5 & | & 6 \end{bmatrix} \] Here, each row corresponds to an equation in the system, and each column represents the coefficients of the variables and the constants. The vertical line separates the coefficients from the constants, indicating the equals sign in the original equations.

• First row: Coefficients \( 1 \) and \( 2 \) for \( x \) and \( y \), with constant \( 3 \).
• Second row: Coefficients \( 4 \) and \( 5 \) for \( x \) and \( y \), with constant \( 6 \).

Using this format, operations can be performed directly on the matrix to solve for the variables.

System of Equations

A system of equations is a set of two or more equations with the same variables. The goal is to find values for the variables that satisfy all of these equations simultaneously.
In the given exercise, we have a system consisting of two linear equations:

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

These equations have the variables \( x \) and \( y \), and what we're looking to find is a single pair \( (x, y) \) that satisfies both equations. Solving such systems often requires methods like substitution, elimination, or using matrices, as in Gaussian elimination.

Row Operations

Row operations are key in manipulating augmented matrices to solve systems of equations through Gaussian elimination. There are three types of row operations:

• Swapping two rows.
• Multiplying a row by a non-zero scalar.
• Adding or subtracting a multiple of one row to another row.

In our exercise, we use row operations to eliminate variables and simplify the matrix. Specifically, to eliminate \( x \) from the second equation. Multiply the entire first row by 4:

• This transformation turns \[ (1, 2, |, 3) \] into \[ (4, 8, |, 12) \].

Then, subtract it from the second row \( (4, 5, |, 6) \):

• The result is a new row: \( (0, -3, |, -6) \), effectively removing \( x \) from the second equation.

Linear Equations

Linear equations form the backbone of systems like the one we solved. A linear equation is an equation that creates a straight line when graphed. It has the general form: \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants, \( x \) and \( y \) are variables.
The solution to a pair of linear equations is where their respective lines intersect on the graph.

• For the equation \( x + 2y = 3 \), if \( x = 0 \) then \( 2y = 3 \), so \( y = \frac{3}{2} \).
• For \( 4x + 5y = 6 \), if \( x = 0 \) then \( 5y = 6 \), so \( y = \frac{6}{5} \).

The entire process of solving these equations via the Gaussian elimination effectively finds the precise point that lies on both lines, giving our solution: \( x = -1 \) and \( y = 2 \). This point is where both linear equations intersect.