Problem 77 - Precalculus

Question

Solve the system of linear equations using Gauss-Jordan elimination. $$\begin{array}{rr} x+y+z= & 3 \\ x-z= & 1 \\ y-z= & -4 \end{array}$$

Step-by-Step Solution

Answer

The solutions are \(x = 3\), \(y = -2\), \(z = 2\).

Step 1: Write the coefficient matrix

Start by writing down the augmented matrix for the system of equations given:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \1 & 0 & -1 & | & 1 \0 & 1 & -1 & | & -4\end{bmatrix}\]

Step 2: Achieve the Upper Triangular Form

Use row operations to transform the matrix to row-echelon form. Subtract Row 1 from Row 2 to make the element below the leading 1 in column 1 zero:\[R_2 = R_2 - R_1 \]The matrix becomes:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0 & -1 & -2 & | & -2 \0 & 1 & -1 & | & -4\end{bmatrix}\]

Step 3: Continue Row Operations

Next, add Row 2 to Row 3 to make the first element of Row 3 zero:\[R_3 = R_3 + R_2 \]The matrix now is:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0 & -1 & -2 & | & -2 \0 & 0 & -3 & | & -6\end{bmatrix}\]

Step 4: Make the leading coefficients 1

Divide Row 2 and Row 3 by their leading coefficients to make them 1:\[R_2 = -R_2, \ R_3 = \frac{1}{-3}R_3\]Resulting in:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0 & 1 & 2 & | & 2 \0 & 0 & 1 & | & 2\end{bmatrix}\]

Step 5: Achieve Reduced Row Echelon Form

Eliminate the non-zero elements above each leading 1 by adjusting Row 1 and Row 2. Subtract the third row from the first and second row to clear the third column:\[R_1 = R_1 - R_3, \ R_2 = R_2 - 2R_3\]The matrix now becomes:\[\begin{bmatrix}1 & 1 & 0 & | & 1 \0 & 1 & 0 & | & -2 \0 & 0 & 1 & | & 2\end{bmatrix}\]

Step 6: Solve for variables

Finally, subtract Row 2 from Row 1 to clear the second column:\[R_1 = R_1 - R_2\]The complete reduced row-echelon form is:\[\begin{bmatrix}1 & 0 & 0 & | & 3 \0 & 1 & 0 & | & -2 \0 & 0 & 1 & | & 2\end{bmatrix}\]This gives the solutions: \(x = 3\), \(y = -2\), \(z = 2\).

Key Concepts

System of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. In our exercise, we have three variables: \(x\), \(y\), and \(z\). The equations are:

• \(x + y + z = 3\)
• \(x - z = 1\)
• \(y - z = -4\)

Solving this system means finding the values of \(x\), \(y\), and \(z\) that satisfy all three equations simultaneously. When lined up, each equation represents a plane in three-dimensional space, and the solution point is where all three planes intersect.
Understanding these intersections helps us visualize why a system can have a single solution, no solution, or infinitely many solutions. In this exercise, using Gauss-Jordan elimination will help to logically work through these intersections and find the specific point of intersection.

Matrix Row Operations

Matrix row operations are the fundamental steps used in the method of Gauss-Jordan elimination to solve systems of linear equations. These operations include:

• Row swapping: Interchanging two rows.
• Row multiplication: Multiplying all terms in a row by a non-zero constant.
• Row addition/subtraction: Adding or subtracting the terms of one row by another.

These operations are key because they help transform the augmented matrix into a form where reading off solutions becomes straightforward.
In our problem, row operations gradually modified the original matrix into reduced row-echelon form, where each leading term in a row is 1 and each column containing a leading 1 has zeros in other rows.
These operations are the building blocks of solving equations, and understanding them is crucial to working with matrices effectively.

Row-Echelon Form

Row-echelon form is a step toward simplifying a system of equations into a more convenient format for solving. A matrix in row-echelon form has all zero rows moving toward the bottom, and each row has its leading coefficient, or 'pivot', to the right of the leading coefficient of the row above. Rows are ordered progressively.
In our problem, we used row operations to arrange the augmented matrix into a step-by-step configuration known as row-echelon form:

• Zeros are introduced below each pivot position.
• Leading coefficients move diagonally to the right as you move down rows.

This format simplifies solving by back-substitution, as it clears away complexities, allowing you to pinpoint the values of each variable efficiently.
Ultimately, achieving row-echelon form is an intermediate step before bringing the matrix to reduced row-echelon form, which provides the direct solutions to the values of the system's variables.