Problem 21 - College Algebra with Corequisite Support

Question

For the following exercises, solve each system by Gaussian elimination. $$ \begin{aligned} 2 x+3 y-4 z &=5 \\ -3 x+2 y+z &=11 \\ -x+5 y+3 z &=4 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \((x, y, z) = \left(\frac{28}{29}, \frac{79}{29}, -\frac{6}{29}\right)\).

Step 1: Write the Augmented Matrix

The given system of equations can be written in matrix form as an augmented matrix. Here, each row of the matrix corresponds to an equation, with the coefficients of the variables and the constants forming the last column: \[\begin{bmatrix} 2 & 3 & -4 & | & 5 \ -3 & 2 & 1 & | & 11 \ -1 & 5 & 3 & | & 4 \end{bmatrix}\]

Step 2: Perform Row Operations to Get Upper Triangular Form

We aim to create zeros below the first element of the first column (pivot element in the first row). We can add appropriate multiples of the first row to other rows:1. Multiply the first row by \(\frac{3}{2}\) and add to the second row.2. Multiply the first row by \(\frac{1}{2}\) and add to the third row.After these operations, the matrix becomes:\[\begin{bmatrix} 2 & 3 & -4 & | & 5 \ 0 & \frac{13}{2} & -\frac{5}{2} & | & \frac{43}{2} \ 0 & \frac{11}{2} & 1 & | & \frac{13}{2} \end{bmatrix}\]

Step 3: Simplify the Matrix Further

Focus on eliminating the first entry of the third row under the second column (which is already zero here). Next, make the second row's first column zero by subtracting appropriate multiples of the second row from the third row:1. Subtract row 2 multiplied by \(\frac{11}{13}\) from row 3.After simplification, we get:\[\begin{bmatrix} 2 & 3 & -4 & | & 5 \ 0 & \frac{13}{2} & -\frac{5}{2} & | & \frac{43}{2} \ 0 & 0 & \frac{29}{13} & | & \frac{-6}{13} \end{bmatrix}\]

Step 4: Back Substitution

We now perform back substitution starting from the last row upwards:1. From the last row: \(\frac{29}{13}z = \frac{-6}{13}\), solve for \(z\): \(z = -\frac{6}{29}\).2. Substitute \(z\) into the second row: \( \frac{13}{2}y - \frac{5}{2}\left(-\frac{6}{29}\right) = \frac{43}{2} \), solve for \(y\).3. Substitute both \(y\) and \(z\) into the first row to solve for \(x\).

Step 5: Solve for All Variables

Solve for each variable explicitly:1. \(z = -\frac{6}{29}\).2. Substitute the value of \(z\) into the second row, calculating \(y = \frac{43 imes 29 + 30}{2 imes 29}\). 3. Substitute values of \(y\) and \(z\) into the first row to find \(x = (5 - 3y + 4z) / 2\).Solve these calculations to find exact values for \(x\), \(y\), and \(z\). Results: \(x = 28/29\), \(y = 79/29\), \(z = -6/29\).

Key Concepts

Augmented Matrix

An augmented matrix is a powerful tool in solving systems of linear equations. It includes the coefficients of the variables in the equations and the constants on the other side of the equations in a single matrix. This format simplifies the manipulation and solution of the equations using methods such as Gaussian elimination.

In the context of our problem, the augmented matrix is formed by taking the coefficients of the variables from each equation and aligning them in a row, with the constants forming the last column. For example, the system is represented by:

• Equation 1: Coefficients for \( x = 2 \), \( y = 3 \), \( z = -4 \), with constant \( 5 \).
• Equation 2: Coefficients for \( x = -3 \), \( y = 2 \), \( z = 1 \), with constant \( 11 \).
• Equation 3: Coefficients for \( x = -1 \), \( y = 5 \), \( z = 3 \), with constant \( 4 \).

So, as an augmented matrix, this becomes:\[ \begin{bmatrix} 2 & 3 & -4 & | & 5 \-3 & 2 & 1 & | & 11 \-1 & 5 & 3 & | & 4 \end{bmatrix} \]This representation helps to apply row operations to simplify and solve the system of equations systematically.

Back Substitution

Back substitution is a method used in Gaussian elimination to solve a system of linear equations after it has been converted into an upper triangular matrix. Continued from our previous discussion on row operations, once the augmented matrix is simplified to an upper triangular form, the variables can be solved starting from the bottom row and working upwards.

Here's how back substitution works:

• Start with the last equation, which typically contains one variable. Solve this equation for that variable.
• Using the known value of this variable, substitute back into the previous equations to solve for the other variables one by one.
• Continue this process until all variables have been solved.

In our example, the simplified matrix looked like:\[ \begin{bmatrix} 2 & 3 & -4 & | & 5 \0 & \frac{13}{2} & -\frac{5}{2} & | & \frac{43}{2} \0 & 0 & \frac{29}{13} & | & \frac{-6}{13} \end{bmatrix} \]Starting from the last row, we can easily find \( z \) and work upwards to find \( y \) and \( x \), ultimately giving us the solution to the system of equations.

System of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. Solving these systems involves finding the values of the variables that satisfy all the equations simultaneously.

Every matrix form of linear equations corresponds to algebraic equations; these can represent anything from a simple intersection between two lines to more complex scenarios in multiple dimensions for multiple equations.

For example, the system given by:\[\begin{aligned} 2x + 3y - 4z &= 5 \-3x + 2y + z &= 11 \-x + 5y + 3z &= 4 \end{aligned} \]represents three equations with variables \( x, y, \) and \( z \). The goal is to find the values of these variables that make all three equations true at the same time.

• This can be visually thought of as finding the intersection point of three planes in space defined by each equation.
• Using methods like Gaussian elimination, these can be systematically transformed into a simpler form, making the job of solving them much easier.
• Ultimately, solving a system of linear equations can help us understand relationships between variables in a multitude of real-world applications, from physics to economics.