Problem 24 - College Algebra with Corequisite Support

Question

For the following exercises, solve each system by Gaussian elimination. $$ \begin{array}{l} 5 x-3 y+4 z=-1 \\ -4 x+2 y-3 z=0 \\ -x+5 y+7 z=-11 \end{array} $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{11}{37}\), \(y = \frac{62}{37}\), \(z = \frac{-24}{37}\).

Step 1: Write the System as an Augmented Matrix

The given system of equations can be transformed into an augmented matrix as follows:\[\begin{bmatrix} 5 & -3 & 4 & | & -1 \-4 & 2 & -3 & | & 0 \-1 & 5 & 7 & | & -11\end{bmatrix}\]

Step 2: Perform Row Operations to Achieve Upper Triangular Form

We start by using the first row to eliminate the first coefficient in the second and third rows.- Replace Row 2 (R2) with \((4 \times R1) + R2\):\[\begin{bmatrix} 5 & -3 & 4 & | & -1 \0 & -10 & 13 & | & -4 \-1 & 5 & 7 & | & -11\end{bmatrix}\]- Replace Row 3 (R3) with \((R1 + 5 \times R3)\):\[\begin{bmatrix} 5 & -3 & 4 & | & -1 \0 & -10 & 13 & | & -4 \0 & 20 & 11 & | & -16\end{bmatrix}\]

Step 3: Normalize the Second Row and Eliminate Second Variable from Third Row

Divide Row 2 by -10 to simplify the second row:\[\begin{bmatrix} 5 & -3 & 4 & | & -1 \0 & 1 & -1.3 & | & 0.4 \0 & 20 & 11 & | & -16\end{bmatrix}\]- Use the new Row 2 to eliminate the second variable in Row 3: Replace Row 3 with \((-20 \times R2) + R3\):\[\begin{bmatrix} 5 & -3 & 4 & | & -1 \0 & 1 & -1.3 & | & 0.4 \0 & 0 & 37 & | & -24\end{bmatrix}\]

Step 4: Back Substitution to Solve for Variables

Starting from the bottom, solve for each variable:1. From Row 3, find \(z\): \[37z = -24 \implies z = \frac{-24}{37}\]2. Substitute \(z\) into Row 2 to solve for \(y\): \[y - 1.3\left(\frac{-24}{37}\right) = 0.4 \implies y = \left(0.4 + \frac{31.2}{37}\right) = 1.243giving y = \frac{62}{37}\]3. Substitute both \(y\) and \(z\) into Row 1 to solve for \(x\): \[5x - 3\left(\frac{62}{37}\right) + 4\left(\frac{-24}{37}\right) = -1 \implies x = \frac{11}{37}\]

Step 5: Conclusion with Final Solutions

The solution to the system of equations after applying Gaussian elimination and back substitution is:\[\begin{align*}x &= \frac{11}{37} \y &= \frac{62}{37} \z &= \frac{-24}{37}\end{align*}\]

Key Concepts

Augmented Matrix

An augmented matrix is a compact representation of a system of linear equations. It combines the coefficients of the variables and the constants of the equations into a single matrix format. For example, consider this system of equations:

• 5x - 3y + 4z = -1
• -4x + 2y - 3z = 0
• -x + 5y + 7z = -11

To form the augmented matrix, arrange the coefficients of each variable into rows. Place the constant term on the right side, separated by a line. For our system, the augmented matrix looks like this:\[\begin{bmatrix}5 & -3 & 4 & | & -1 \-4 & 2 & -3 & | & 0 \-1 & 5 & 7 & | & -11\end{bmatrix}\]This format helps us perform systematic operations to solve the system using Gaussian elimination.

Upper Triangular Form

The upper triangular form is a matrix form where all elements below the main diagonal are zero. Achieving this form is a crucial step in solving systems of linear equations via Gaussian elimination.
To transform a matrix to an upper triangular form, you perform row operations to create zeros below the leading coefficient of the first row, generally known as the pivot. This process continues for each pivot in subsequent rows. Here is an example with our augmented matrix:\[\begin{bmatrix}5 & -3 & 4 & | & -1 \0 & -10 & 13 & | & -4 \0 & 20 & 11 & | & -16\end{bmatrix}\]You can see that the positions below the leading coefficients in each row are zero. This form simplifies the process of back substitution, making it easier to solve for individual variables.

Back Substitution

Once the matrix is in upper triangular form, back substitution is the method used to solve for the variables. Starting from the bottom row, which usually contains only one variable due to prior row operations, you can easily solve it.
In our example, the last row gives the equation for z:\[37z = -24\]This lets you solve directly for z. After solving for one variable, you substitute it into the previous row to find the next variable. For instance:

• Solve for z.
• Substitute z into the second row to find y.
• Substitute both y and z into the first row to find x.

Continuing these steps ensures that each variable is solved in terms of those that were already determined, resulting in the full solution to the system: \[x = \frac{11}{37}, \; y = \frac{62}{37}, \; z = \frac{-24}{37}\]

Row Operations

Row operations are fundamental in transforming a matrix during Gaussian elimination. They allow you to manipulate rows to create zeros in specific positions, aiding the transition to upper triangular form. There are three primary 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 problem, these operations are applied to eliminate coefficients step-by-step. For example, transforming the second row might involve multiplying the first row by a constant and adding it to the second row:\[(4 \times \text{Row 1}) + \text{Row 2} = -4x + 2y - 3z = 0\]This operation results in a row where the first element, once a coefficient of x, becomes zero. Applying such operations strategically reduces the matrix to upper triangular form, preparing it for back substitution.