Problem 26 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system by Gaussian elimination. $$ \begin{array}{l} 3 x+4 y=12 \\ -6 x-8 y=-24 \end{array} $$

Step-by-Step Solution

Answer

The system has infinitely many solutions: \( x = \frac{12 - 4t}{3}, y = t \) for any real number \( t \).

Step 1: Write the system as an augmented matrix

First, convert the given system of equations into an augmented matrix. The system is: \[ 3x + 4y = 12 \] \[ -6x - 8y = -24 \] So the augmented matrix is: \[ \begin{bmatrix} 3 & 4 & | & 12 \ -6 & -8 & | & -24 \end{bmatrix} \]

Step 2: Apply row operations to simplify

We want to transform the matrix into row-echelon form. Start by adding twice the first row to the second row to create a zero in the first column of the second row: Row 2 <- Row 2 + 2 * Row 1 \[ \begin{bmatrix} 3 & 4 & | & 12 \ 0 & 0 & | & 0 \end{bmatrix} \]

Step 3: Analyze the resulting matrix

The resulting matrix is transformed into: \[ \begin{bmatrix} 3 & 4 & | & 12 \ 0 & 0 & | & 0 \end{bmatrix} \] The second row indicates that 0 = 0, which is always true, meaning the system does not provide additional information.

Step 4: Conclude with the solution of the system

The system has a rank of 1 which is less than the number of variables. This means the system has infinitely many solutions. Expressing the solution in terms of a parameter 't', the first equation becomes:\[ 3x + 4y = 12 \] Solve for y: \[ y = t \] \[ x = \frac{12 - 4t}{3} \] Thus, the solution is: \[ x = \frac{12 - 4t}{3}, \ y = t \] where \( t \) is any real number.

Key Concepts

Systems of Linear Equations

A system of linear equations is a set of equations with multiple variables. Each equation is linear, meaning the variables are raised only to the power of one. Examples include two-variable systems like \(3x + 4y = 12\) and \(-6x - 8y = -24\). The goal is often to find values for the variables that satisfy all equations simultaneously. These systems can have a unique solution, no solution, or infinitely many solutions, depending on the equations' consistency and independence. Solving these systems helps in various applications, from engineering to economics.

One effective method to tackle these systems is Gaussian elimination, where equations are manipulated to find solutions systematically. This process involves transforming the system into a form that's easier to solve, usually aiming for one equation per variable.

Augmented Matrix

An augmented matrix is a compact way of representing a system of linear equations. It combines the coefficients of the variables and the constants from the equations into a single array. For example, consider the system:

• \(3x + 4y = 12\)
• \(-6x - 8y = -24\)

The augmented matrix form would be:\[\begin{bmatrix}3 & 4 & | & 12 \-6 & -8 & | & -24\end{bmatrix}\]Here, the vertical bar separates the coefficients from the constants. This format is instrumental in Gaussian elimination as it sets the stage for applying row operations without needing to constantly rewrite equations. Converting a system to an augmented matrix is the first step towards solving it algebraically using matrices.

Row Operations

Row operations are essential manipulations performed on an augmented matrix to simplify a system of linear equations. These operations include:

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

In our example, we added twice the first row to the second row to make a zero appear in the second row's first column. This simplification steps up the matrix towards a form called row-echelon, or eventually, reduced row-echelon form. The aim is to isolate each variable step-by-step, making calculations and seeing relationships between equations more transparent.

Infinite Solutions

When solving systems of equations, a scenario arises where you derive an equation like "0 = 0" from one of the matrix rows after some row operations. This indicates that one of the original equations was a linear combination of others, leading to dependency. Thus, we're left with a situation of infinitely many solutions.

In the transformed matrix:\[\begin{bmatrix}3 & 4 & | & 12 \0 & 0 & | & 0\end{bmatrix}\]The first row still has two variables, and no unique solution emerges directly from the second row. Instead, we express one variable in terms of another using a parameter like \(t\). For the example:

• \(y = t\)
• \(x = \frac{12 - 4t}{3}\)

Here, \(t\) can be any real number, highlighting the endless solution possibilities that fit the original equations.