Problem 25 - Advanced Engineering Mathematics

Question

In Problems 25 and 26, solve the given system of equations by Gauss-Jordan elimination. $$ \left(\begin{array}{rrr} 5 & -1 & 1 \\ 2 & 4 & 0 \\ 1 & 1 & 5 \end{array}\right) \mathbf{X}=\left(\begin{array}{r} -9 \\ 27 \\ 9 \end{array}\right) $$

Step-by-Step Solution

Answer

\(x_1 = -\frac{41}{5}, x_2 = 4, x_3 = \frac{9}{10}\)

Step 1: Write the augmented matrix

First, we need to write the augmented matrix for the given system of equations. The augmented matrix combines the coefficients from the left side with the constants from the right side: \[\left(\begin{array}{ccc|c}5 & -1 & 1 & -9 \2 & 4 & 0 & 27 \1 & 1 & 5 & 9\end{array}\right)\]

Step 2: Create a leading 1 in the first row

To begin the process of Gauss-Jordan elimination, the first leading coefficient should be 1. Divide the entire first row by 5 to achieve this leading 1:\[\left(\begin{array}{ccc|c}1 & -\frac{1}{5} & \frac{1}{5} & -\frac{9}{5} \2 & 4 & 0 & 27 \1 & 1 & 5 & 9\end{array}\right)\]

Step 3: Eliminate first column entries beneath the leading 1

Subtract 2 times the first row from the second row and subtract the first row from the third row to create zeros below the first leading 1:\[\left(\begin{array}{ccc|c}1 & -\frac{1}{5} & \frac{1}{5} & -\frac{9}{5} \0 & \frac{22}{5} & -\frac{2}{5} & \frac{81}{5} \0 & \frac{6}{5} & \frac{24}{5} & \frac{54}{5}\end{array}\right)\]

Step 4: Make a leading 1 in the second row

In order to simplify further, we create a leading 1 in the second row, second column by dividing the second row by \(\frac{22}{5}\) (which simplifies to multiplying by \(\frac{5}{22}\)):\[\left(\begin{array}{ccc|c}1 & -\frac{1}{5} & \frac{1}{5} & -\frac{9}{5} \0 & 1 & -\frac{1}{11} & \frac{81}{22} \0 & \frac{6}{5} & \frac{24}{5} & \frac{54}{5}\end{array}\right)\]

Step 5: Eliminate second column entries above and below the leading 1

Eliminate the second column entries by adding \(\frac{1}{5}\) times the second row to the first row and subtracting \(\frac{6}{5}\) times the second row from the third row:\[\left(\begin{array}{ccc|c}1 & 0 & \frac{2}{11} & -\frac{103}{22} \0 & 1 & -\frac{1}{11} & \frac{81}{22} \0 & 0 & \frac{30}{11} & \frac{27}{11}\end{array}\right)\]

Step 6: Make a leading 1 in the third row

Create a leading 1 in the third row by dividing the third row by \(\frac{30}{11}\) (which simplifies to multiplying by \(\frac{11}{30}\)):\[\left(\begin{array}{ccc|c}1 & 0 & \frac{2}{11} & -\frac{103}{22} \0 & 1 & -\frac{1}{11} & \frac{81}{22} \0 & 0 & 1 & \frac{9}{10}\end{array}\right)\]

Step 7: Eliminate third column entries above the leading 1

Finally, eliminate the third column entries above the leading 1 by subtracting \(\frac{2}{11}\) times the third row from the first row and adding \(\frac{1}{11}\) times the third row to the second row:\[\left(\begin{array}{ccc|c}1 & 0 & 0 & -\frac{41}{5} \0 & 1 & 0 & 4 \0 & 0 & 1 & \frac{9}{10}\end{array}\right)\]

Step 8: Interpret the solution from the matrix

The resultant matrix is in reduced row echelon form, and the solutions are directly visible from the last column: \[X = \left(\begin{array}{c}x_1 \x_2 \x_3\end{array}\right) = \left(\begin{array}{c}-\frac{41}{5} \4 \\frac{9}{10}\end{array}\right)\]

Key Concepts

Augmented Matrix

The augmented matrix is a critical component in solving systems of equations, especially when employing techniques such as Gauss-Jordan elimination. It combines both the coefficients of the variables and the constants from each equation into a single matrix format. This is done by appending the constants from the equations as an additional column to the coefficient matrix. For the given system of equations, the augmented matrix appears as follows:

• The left side contains the coefficients of the variables arranged in rows and columns.
• A vertical bar separates these coefficients from the constants, which form the last column.

To form the augmented matrix, each equation in the system corresponds to a row in the matrix. This crucial step allows us to visualize and manipulate the system in matrix form, paving the way for the elimination processes to follow. By converting equations to an augmented matrix, we harness the power of matrix operations, making the solution process structured and systematic.

Leading Coefficient

In matrix transformations, the leading coefficient is the first non-zero number in any row, from left to right. It's pivotal because it's used to simplify the matrix into its row-echelon form and ultimately its reduced row-echelon form. Finding a leading coefficient of 1 in a row is essential in Gauss-Jordan elimination, as it helps set the stage for eliminating other numbers in that column through simple arithmetic operations.

• To create a leading 1, the row can be scaled (divided by its leading entry).
• Once a leading 1 is established in the first column of the row, other entries in the same column but different rows must be zeroed out using row operations.

This process is repeated for subsequent columns, always working from the top row downward and from left to right. The resulting matrix will have a "stair-step" pattern, which makes it easier to work downwards towards a solution. Understanding the importance of leading coefficients is key to mastering matrix manipulations.

Reduced Row Echelon Form

The reduced row echelon form (RREF) is the final configuration we aim for in the process of solving systems of equations using Gauss-Jordan elimination. A matrix is in RREF if it meets specific criteria:

• Every leading (nonzero) entry in a row is 1.
• Each leading 1 is in a column that has zeros in all its other entries.
• The leading 1 in each row is to the right of any leading 1s in the rows above it.
• Any zero rows (rows where every entry is zero) are at the bottom of the matrix.

Achieving RREF means that the matrix represents a system of equations that is straightforward to solve. The solutions can be read directly from the matrix since each leading 1 represents a distinct variable and other entries in the last column give the solution for each. Reaching RREF systematically eliminates ambiguous situations and clarifies the solutions, ensuring precise and accurate algebraic results.