Problem 36 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system by Gaussian elimination. $$ \left[\begin{array}{rrr|r} -0.1 & 0.3 & -0.1 & 0.2 \\ -0.4 & 0.2 & 0.1 & 0.8 \\ 0.6 & 0.1 & 0.7 & -0.8 \end{array}\right] $$

Step-by-Step Solution

Answer

\(x_1 = 0.2857, x_2 = 0.7619, x_3 = -1.5238\)

Step 1: Write the Augmented Matrix

The given system is already in the form of an augmented matrix: \[ \left[\begin{array}{rrr|r} -0.1 & 0.3 & -0.1 & 0.2 \ -0.4 & 0.2 & 0.1 & 0.8 \ 0.6 & 0.1 & 0.7 & -0.8 \end{array}\right] \] Our goal is to transform this matrix into an upper triangular form using Gaussian elimination.

Step 2: Make Pivot in Row 1

To create a 1 leading the row, we first multiply Row 1 by -10 to eliminate the decimal numbers: \[ R_1 = -10 \times R_1: \left[ \begin{array}{rrr|r} 1 & -3 & 1 & -2 \end{array} \right] \] Now the matrix becomes: \[ \left[\begin{array}{rrr|r} 1 & -3 & 1 & -2 \ -0.4 & 0.2 & 0.1 & 0.8 \ 0.6 & 0.1 & 0.7 & -0.8 \end{array}\right] \]

Step 3: Eliminate Below Pivot in Row 1

To eliminate the coefficients below the pivot in Column 1, perform the following row operations: \[ R_2 = R_2 + 0.4 \times R_1 \] \[ R_3 = R_3 - 0.6 \times R_1 \] Applying these, we get: \[ \left[\begin{array}{rrr|r} 1 & -3 & 1 & -2 \ 0 & -1 & 0.5 & 0 \ 0 & 1.9 & 0.1 & -1.6 \end{array}\right] \]

Step 4: Make Pivot in Row 2

To make the pivot a 1 in Row 2, multiply Row 2 by -1:\[ R_2 = -1 \times R_2: \left[ \begin{array}{rrr|r} 0 & 1 & -0.5 & 0 \end{array} \right] \] Now the matrix becomes: \[ \left[\begin{array}{rrr|r} 1 & -3 & 1 & -2 \ 0 & 1 & -0.5 & 0 \ 0 & 1.9 & 0.1 & -1.6 \end{array}\right] \]

Step 5: Eliminate Below Pivot in Row 2

To eliminate the coefficients below the pivot in Column 2, perform the row operation on Row 3: \[ R_3 = R_3 - 1.9 \times R_2 \] This changes the matrix to: \[ \left[\begin{array}{rrr|r} 1 & -3 & 1 & -2 \ 0 & 1 & -0.5 & 0 \ 0 & 0 & 1.05 & -1.6 \end{array}\right] \]

Step 6: Make Pivot in Row 3

Divide Row 3 by 1.05 to turn the pivot element into 1: \[ R_3 = \frac{1}{1.05} \cdot R_3: \left[ \begin{array}{rrr|r} 0 & 0 & 1 & -\frac{1.6}{1.05} \end{array} \right] \] Now the matrix becomes:\[ \left[\begin{array}{rrr|r} 1 & -3 & 1 & -2 \ 0 & 1 & -0.5 & 0 \ 0 & 0 & 1 & -1.5238 \end{array}\right] \]

Step 7: Back Substitution for x3

We now have an upper triangular matrix. From the third row, we read: \[ x_3 = -1.5238 \]

Step 8: Back Substitution for x2

Using the second row, substitute the value of \(x_3\) to get \(x_2\): \[ 0.5 \times (-1.5238) - 0.5 \times x_3 = 0 \]\[ x_2 - 0.5(-1.5238) = 0 \]\[ x_2 = 0.7619 \]

Step 9: Back Substitution for x1

Using the first row, substitute the values of \(x_2\) and \(x_3\) to solve for \(x_1\): \[ 1 - 3(0.7619) + (-1.5238) = -2 \] Let's compute: \[ x_1 = -2 + 3(0.7619) - 1.5238 \]\[ x_1 = 0.2857 \]

Key Concepts

Augmented Matrix

An augmented matrix is a very handy tool to use when solving systems of linear equations. It combines the coefficients and the constants of the equations into one matrix. This matrix makes it easier to perform operations to solve the system. For example, when you see a system like the one given in the exercise, you can write it as an augmented matrix:

• The first part to the left of the vertical line (|) includes all the coefficients of the unknowns.
• The part to the right of the vertical line contains the constants from each equation.

This structure helps visualize the problem, as it allows one to apply operations to rows in a straightforward manner. The aim is to use these operations to simplify the matrix into a more usable form, like an upper triangular matrix.

Upper Triangular Matrix

An upper triangular matrix is a key concept when using Gaussian elimination. This type of matrix has all its elements equal to zero below the main diagonal. For instance, if you have a 3x3 matrix, the positions (2,1) and (3,1), (3,2) should be zeros for it to be upper triangular.

• The process of transforming your augmented matrix into this form is called row reduction.
• The goal is to clear out the lower triangle of the matrix so you end up with a staircase-like structure.
• Once you have an upper triangular matrix, it becomes much easier to solve for each unknown variable using back substitution.

Obtaining an upper triangular matrix simplifies the system of equations, making the solution significantly faster and more straightforward.

Back Substitution

Back substitution is the final step in solving a system of equations using Gaussian elimination. Once you've achieved an upper triangular matrix, back substitution allows you to find the values of the variables step by step.

• Start from the bottom row and solve for the last unknown variable. This usually involves a simple equation with only one variable.
• Work your way up the rows, using the value of the variable from the row below to find the next unknown.
• Continue this process until you've determined all the variable values.

Back substitution allows for a logical and efficient way to arrive at the solution in a structured manner. For example, the third row gives you the exact value of one variable, which helps to substitute upwards until all variables have a defined value.

Pivot Element

The pivot element is a very important part of Gaussian elimination. It is the element in the row that is used as a reference for the elimination process.

• A pivot is ideally the number '1', which simplifies calculations, or it can be made '1' by row operations.
• Each pivot should be the leading 1 in its row, meaning it is the first non-zero entry from the left in its row.
• Choosing and manipulating pivots correctly allows you to clear the columns below them to zeros more easily.

In the solution, pivots are established through strategic row operations, ensuring the matrix becomes upper triangular while maintaining the system's balance. Handling pivots correctly is crucial for simplifying matrices and is a skill that makes solving complex systems more manageable.