Problem 75 - Precalculus

Question

Solve the system of linear equations using Gauss-Jordan elimination. $$\begin{array}{rr} x-2 y+3 z= & 5 \\ 3 x+6 y-4 z= & -12 \\ -x-4 y+6 z= & 16 \end{array}$$

Step-by-Step Solution

Answer

The solution is \(x = -\frac{17}{6}, y = -\frac{3}{2}, z = \frac{14}{3}\).

Step 1: Write the augmented matrix

The first step is to represent the given system of linear equations in an augmented matrix form. The system is: \[\begin{align*}x - 2y + 3z &= 5 \3x + 6y - 4z &= -12 \-x - 4y + 6z &= 16\end{align*}\]The augmented matrix is:\[\begin{bmatrix}1 & -2 & 3 & | & 5 \3 & 6 & -4 & | & -12 \-1 & -4 & 6 & | & 16\end{bmatrix}\]

Step 2: Perform row operations to get a leading 1 in the first column

Pivot the matrix to have 1 in the first position of the first row (this is already done). Use the first row to make the other elements in the first column zero by performing the following operations:- Replace \(R_2\) with \(R_2 - 3R_1\)- Replace \(R_3\) with \(R_3 + R_1\)Performing these operations, we get:\[\begin{bmatrix}1 & -2 & 3 & | & 5 \0 & 12 & -13 & | & -27 \0 & -6 & 9 & | & 21\end{bmatrix}\]

Step 3: Making the next pivot a leading 1

Now, we need to make the pivot in the second row, second column a leading 1. We can divide the entire second row by 12:Replace \(R_2\) with \(\frac{1}{12}R_2\)\[\begin{bmatrix}1 & -2 & 3 & | & 5 \0 & 1 & -\frac{13}{12} & | & -\frac{27}{12} \0 & -6 & 9 & | & 21\end{bmatrix}\]

Step 4: Make all other elements in the second column zero

Using the second row, make the elements above and below the pivot in the second column zeros by performing the following operations:- Replace \(R_3\) with \(R_3 + 6R_2\)- Replace \(R_1\) with \(R_1 + 2R_2\)Resulting matrix:\[\begin{bmatrix}1 & 0 & \frac{11}{6} & | & \frac{1}{2} \0 & 1 & -\frac{13}{12} & | & -\frac{27}{12} \0 & 0 & \frac{3}{2} & | & 7\end{bmatrix}\]

Step 5: Normalize the third row

Make the pivot in the third row, third column a leading 1 by replacing \(R_3\) with \(\frac{2}{3}R_3\):\[\begin{bmatrix}1 & 0 & \frac{11}{6} & | & \frac{1}{2} \0 & 1 & -\frac{13}{12} & | & -\frac{27}{12} \0 & 0 & 1 & | & \frac{14}{3}\end{bmatrix}\]

Step 6: Make all other elements in the third column zero

Finally, eliminate all other elements in the third column using the third row by performing:- Replace \(R_1\) with \(R_1 - \frac{11}{6}R_3\)- Replace \(R_2\) with \(R_2 + \frac{13}{12}R_3\)This gives us:\[\begin{bmatrix}1 & 0 & 0 & | & -\frac{17}{6} \0 & 1 & 0 & | & -\frac{3}{2} \0 & 0 & 1 & | & \frac{14}{3}\end{bmatrix}\]

Step 7: Read the solution from the matrix

The matrix now represents the system of equations in reduced row-echelon form:\[\begin{align*}x &= -\frac{17}{6} \y &= -\frac{3}{2} \z &= \frac{14}{3}\end{align*}\]Hence, the solution is the set \((x, y, z) = (-\frac{17}{6}, -\frac{3}{2}, \frac{14}{3})\).

Key Concepts

system of linear equations

A system of linear equations involves finding the values of variables that satisfy multiple linear equations simultaneously. Linear equations consist of terms that are either constants or products of constants and a single variable.
Each equation in the system aligns on a straight line when graphed, and the solution, if it exists, is the point where these lines intersect. For example, the system given is:

• \(x - 2y + 3z = 5\)
• \(3x + 6y - 4z = -12\)
• \(-x - 4y + 6z = 16\)

A solution to this system would be a specific set of values for \(x\), \(y\), and \(z\) such that when substituted into each equation, all the equations hold true. If no such set exists, the system may be inconsistent. If there are multiple solutions, the system is dependent, often representing a series of infinite solutions that lay on the same line or plane. In most cases like this one, you want to find the unique solution that satisfies all equations using methods like Gauss-Jordan elimination.

augmented matrix

An augmented matrix is a compact representation of a system of linear equations, capturing both the coefficients of the variables and the constants on the right side of the equations. This matrix simplifies the solving process by allowing operations to be performed directly on the entire set.To convert a system into an augmented matrix, write the coefficients of each variable and the constant term from each equation in a row. For the given system, the augmented matrix is:\[\begin{bmatrix}1 & -2 & 3 & | & 5 \3 & 6 & -4 & | & -12 \-1 & -4 & 6 & | & 16\end{bmatrix}\]Here, each row aligns with an equation from the system. The vertical line separates the coefficients from the constants, illustrating where operations will aim to achieve reduced row-echelon form. This setup aids in systematic transformations through row operations to reveal the solution.

row operations

Row operations are the core techniques used during the Gauss-Jordan elimination process to manipulate the augmented matrix towards its reduced row-echelon form. By systematically applying these, the matrix can be simplified until the solution to the system of equations is neatly extractable.There are three types of row operations:

• Swapping two rows: Allows repositioning of equations for convenience.
• Multiplying a row by a nonzero scalar: Adjusts the equation's magnitude uniformly without changing the solution.
• Adding or subtracting a multiple of one row to another: Used to eliminate variables and create zeros in specific positions.

In the provided problem solution, these operations transform the original matrix until each row represents a distinct equation with a leading 1 followed by zeros (except the last column, which stores the solution). These steps sequentially make diagonal elements 1 and eliminate other entries in each column, resulting in a matrix form where the variables easily reveal their values: \(x = -\frac{17}{6}\), \(y = -\frac{3}{2}\), and \(z = \frac{14}{3}\).