Problem 40 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the system of linear equations using the Gauss-Jordan elimination method. \(\begin{array}{rr}2 x+y-2 z= & 4 \\ x+3 y-z= & -3 \\ 3 x+4 y-z= & 7\end{array}\)

Step-by-Step Solution

Answer

The short answer to the given system of linear equations using the Gauss-Jordan elimination method is: \(x = 10\), \(y = -2\), and \(z = 7\).

Step 1: Write the augmented matrix

We write the given system of equations as an augmented matrix. The matrix consists of the coefficients of the variables on the left side of each equation and constants on the right side: \[ \left[\begin{array}{rrr|r} 2 & 1 & -2 & 4 \\ 1 & 3 & -1 & -3 \\ 3 & 4 & -1 & 7 \end{array}\right] \]

Step 2: Swap row 1 and row 2

We want a 1 in the top left corner, so we swap row 1 and row 2: \[ \left[\begin{array}{rrr|r} 1 & 3 & -1 & -3 \\ 2 & 1 & -2 & 4 \\ 3 & 4 & -1 & 7 \end{array}\right] \]

Step 3: Eliminate x coefficients in row 2 and row 3

We use row operations to eliminate the x coefficients in row 2 and row 3. We subtract 2 times row 1 from row 2 and 3 times row 1 from row 3: \[ \left[\begin{array}{rrr|r} 1 & 3 & -1 & -3 \\ 0 & -5 & 0 & 10 \\ 0 & -5 & 2 & 16 \end{array}\right] \]

Step 4: Divide row 2 by -5

We divide row 2 by -5 to make the leading coefficient in row 2 equal to 1: \[ \left[\begin{array}{rrr|r} 1 & 3 & -1 & -3 \\ 0 & 1 & 0 & -2 \\ 0 & -5 & 2 & 16 \end{array}\right] \]

Step 5: Eliminate y coefficients in row 1 and row 3

We use row operations to eliminate the y coefficients in row 1 and row 3. We subtract 3 times row 2 from row 1 and add row 2 to row 3: \[ \left[\begin{array}{rrr|r} 1 & 0 & -1 & 3 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 2 & 14 \end{array}\right] \]

Step 6: Divide row 3 by 2

We divide row 3 by 2 to make the leading coefficient in row 3 equal to 1: \[ \left[\begin{array}{rrr|r} 1 & 0 & -1 & 3 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 7 \end{array}\right] \]

Step 7: Eliminate z coefficient in row 1

We use row operations to eliminate the z coefficients in row 1. We add row 3 to row 1: \[ \left[\begin{array}{rrr|r} 1 & 0 & 0 & 10 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 7 \end{array}\right] \]

Step 8: Read the solution

Now our augmented matrix is in reduced row-echelon form. We can read off the solution directly: \(x = 10\), \(y = -2\), and \(z = 7\).

Key Concepts

Linear Equations

Linear equations are mathematical statements that graph as straight lines. They are often in the form \( ax + by + cz = d \), where \( a, b, \) and \( c \) are constants, and \( x, y, \) and \( z \) are variables. Each term is either a constant or the product of a constant and a single variable.
Linear systems like the one in the exercise consist of multiple linear equations. Each equation corresponds to a line in a multidimensional space, where intersections represent potential solutions.
The system provided in the exercise has three equations. This means you are searching for a point where all three lines intersect, revealing the values of the variables \( x, y, \) and \( z \). Solving such systems is crucial in many fields like physics, engineering, and forecasting, helping predict outcomes or determine relationships among variables.

Augmented Matrix

An augmented matrix is a compact way of representing a system of linear equations. It combines the coefficients of the variables and the constants from each equation into a matrix format. This method streamlines solving equations using techniques like Gauss-Jordan elimination.
Consider the system of equations from the exercise. You represent it as an augmented matrix: \[\left[\begin{array}{rrr|r}2 & 1 & -2 & 4 \1 & 3 & -1 & -3 \3 & 4 & -1 & 7\end{array}\right]\] Here, each row corresponds to an equation, and each column contains the coefficients for \( x, y, \) and \( z \), respectively, with the last column representing the constants.
Using an augmented matrix simplifies complex systems of equations, allowing for efficient manipulation through row operations. This form is particularly useful in computational methods and helps visualize the steps in the solution process.

Row Operations

Row operations are essential tools in linear algebra used to simplify matrices, making it easier to achieve a solution. There are three main types of row operations: swapping rows, multiplying a row by a non-zero scalar, and adding or subtracting a multiple of one row to another.
During Gauss-Jordan elimination, these operations help transform an augmented matrix into reduced row-echelon form, where the solution can be easily read.
For example, in the exercise, the row operation steps include:

• Swapping rows to reorder terms for easier solving.
• Eliminating coefficients by subtracting multiples of one row from another, zeroing out elements to simplify equations.
• Dividing rows by coefficients to normalize leading values.

These operations maintain the system's integrity, ensuring that any solution found remains consistent with the original equations. They offer a logical and structured path to unlock the values of unknown variables.