Problem 42 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the system of linear equations using the Gauss-Jordan elimination method. \(\begin{aligned} 2 x+3 y-2 z=& 10 \\ 3 x-2 y+2 z=& 0 \\ 4 x-y+3 z=&-1 \end{aligned}\)

Step-by-Step Solution

Answer

The solution to the given system of linear equations using the Gauss-Jordan elimination method is: \(x = 1, y = 2, z = -\frac{5}{3}\).

Step 1: 1. Create the augmented matrix

\[\left[\begin{array}{ccc|c} 2 & 3 & -2 & 10 \\ 3 & -2 & 2 & 0 \\ 4 & -1 & 3 & -1 \end{array}\right]\] Next, we will perform row operations to create an upper triangular form.

Step 2: 2. Row swap and row addition

Divide the first row by 2 and swap it with the third row. \[\left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & -\frac{1}{2} \\ 3 & -2 & 2 & 0 \\ 2 & 3 & -2 & 10 \end{array}\right]\] Subtract 3 times the first row from the second row and 2 times the first row from the third row. \[\left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & -\frac{1}{2} \\ 0 & -\frac{1}{2} & -\frac{5}{2} & \frac{3}{2} \\ 0 & 4 & -7 & 11 \end{array}\right]\]

Step 3: 3. Row multiplication

Divide the second row by -\(\frac{1}{2}\) and the third row by 4. \[\left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} & -\frac{3}{2} \\ 0 & 1 & -\frac{7}{4} & \frac{11}{4} \end{array}\right]\]

Step 4: 4. Row addition

Subtract the second row from the third row. \[\left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} & -\frac{3}{2} \\ 0 & 0 & -\frac{3}{4} & \frac{5}{4} \end{array}\right]\] Now that we have the upper triangular form, we can find the values of the variables using back substitution.

Step 5: 5. Back substitution

Solve for z in the third row. \(-\frac{3}{4}z = \frac{5}{4} \implies z = -\frac{5}{3}\) Now, we can find the value of y using z. \(y + \frac{5}{2}(-\frac{5}{3}) = -\frac{3}{2} \implies y = 2\) Finally, we can find the value of x using y and z. \(x - \frac{1}{2}(2) + \frac{3}{2}(-\frac{5}{3}) = -\frac{1}{2} \implies x = 1\)

Step 6: 6. Solution

The values for the variables are: \[x = 1, y = 2, z = -\frac{5}{3}\]

Key Concepts

Linear Equations

A system of linear equations consists of two or more linear equations involving the same set of variables. In our given problem, there is a set of three linear equations:- \(2x + 3y - 2z = 10\)- \(3x - 2y + 2z = 0\)- \(4x - y + 3z = -1\)These equations represent three planes in three-dimensional space, where the solution is the point where these planes intersect. To solve such systems efficiently, the Gauss-Jordan elimination method can be applied, transforming the system into a simpler form using an augmented matrix.

Augmented Matrix

An augmented matrix is a compact form of representing a system of linear equations. It combines the coefficients of the variables and the constant terms into a matrix format:- For the equations in our exercise, we initially have: \[ \begin{array}{ccc|c} 2 & 3 & -2 & 10 \ 3 & -2 & 2 & 0 \ 4 & -1 & 3 & -1 \end{array} \]The vertical line separates the coefficient matrix from the constants on the right side of the equations. This matrix helps in organizing the operations needed for simplification and solution of the system. Converting equations into an augmented matrix is the first step in applying row operations to simplify and solve the system.

Row Operations

Row operations consist of three types of manipulations that can be performed on matrices:

• Swapping two rows
• Multiplying a row by a scalar, which is not zero
• Adding or subtracting the multiple of one row from another row

These operations help in transforming the original matrix into a form where it is easier to identify solutions. In our solution process, row operations are applied to gradually simplify the augmented matrix into row-echelon form, starting with operations like dividing, swapping rows, and subtracting multiples of rows to eliminate variable coefficients and simplify the matrix structure.

Back Substitution

Back substitution is the process used to find the values of variables once the matrix has been reduced to an upper triangular form. This means that the system is simplified such that the equations simplify into a ladder of dependencies:- Start with the last row and solve for the single variable (here, \(z\)): - Solve \(-\frac{3}{4}z = \frac{5}{4}\), yielding \(z = -\frac{5}{3}\).- Substitute \(z\) back into the above rows to solve for \(y\) and \(x\): - From the second equation, solve for \(y\): \(y + \frac{5}{2}(-\frac{5}{3}) = -\frac{3}{2}\), yielding \(y = 2\). - Substitute \(y\) and \(z\) into the first equation to solve for \(x\): \(x - \frac{1}{2}(2) + \frac{3}{2}(-\frac{5}{3}) = -\frac{1}{2}\), which gives \(x = 1\).Back substitution is the final step in the Gauss-Jordan method, allowing for precise determination of each variable's value.