Problem 26 - Beginning and Intermediate Algebra

Question

Solve each systen \(3 x+4 y=-6\) \(-x+3 z=1\) \(2 y+3 z=-1\)

Step-by-Step Solution

Answer

The solution to the given system of linear equations is: \( x = -2, y = 0, z = -\frac{1}{3} \).

Step 1: Creating the augmented matrix for the system

Write down the coefficients of the variables x, y, and z and the constants in an augmented matrix form: \[ \begin{bmatrix} 3 & 4 & 0 & | & -6 \\ -1 & 0 & 3 & | & 1 \\ 0 & 2 & 3 & | & -1 \end{bmatrix} \]

Step 2: Apply Gaussian Elimination to obtain row-echelon form

Perform row operations on the matrix to obtain a row-echelon form: Swap row_1 and row_2 (to have a pivot '1' in the first row): \[ \begin{bmatrix} -1 & 0 & 3 & | & 1 \\ 3 & 4 & 0 & | & -6 \\ 0 & 2 & 3 & | & -1 \end{bmatrix} \] Multiply row_1 by -3 and add it to row_2: \[ \begin{bmatrix} -1 & 0 & 3 & | & 1 \\ 0 & 4 & 9 & | & -3 \\ 0 & 2 & 3 & | & -1 \end{bmatrix} \] Multiply row_3 by 2 and subtract row_2: \[ \begin{bmatrix} -1 & 0 & 3 & | & 1 \\ 0 & 4 & 9 & | & -3 \\ 0 & 0 & -3 & | & 1 \end{bmatrix} \]

Step 3: Solve for variables x, y, and z

From the obtained row-echelon form, we have the following system of equations: \[ \begin{cases} -1x + 0y + 3z = 1 \\ 0x + 4y + 9z = -3 \\ 0x + 0y - 3z = 1 \end{cases} \] Now, we can use back substitution to find the variables: Divide the third equation by -3: \( z = -\frac{1}{3} \) Substitute the value of z into the second equation and solve for y: \( 4y + 9(-\frac{1}{3}) = -3 \) \( 4y - 3 = -3 \) \( y = 0 \) Finally, substitute the values of y and z into the first equation and solve for x: \( -x + 3(-\frac{1}{3}) = 1\) \( -x - 1 = 1 \) \( x = -2 \) The solution to the given system of linear equations is: \( x = -2, y = 0, z = -\frac{1}{3} \).

Key Concepts

Augmented Matrix

When solving a system of linear equations using Gaussian Elimination, the first step is to express the system in an augmented matrix form. An augmented matrix is a compact and efficient way to represent a system of linear equations. It consists of both the coefficients of the variables and the constants from the right-hand side of the equations.

Consider the system of equations given:

• Equation 1: \(3x + 4y = -6\)
• Equation 2: \(-x + 3z = 1\)
• Equation 3: \(2y + 3z = -1\)

To create an augmented matrix, we strip each equation of its variables, keeping only the coefficients, and line them up in rows. The constants form an extra column, separated by a vertical bar to show they're on the equations' right side:

• Row 1 (Equation 1): Coefficients of \(x\), \(y\), \(z\): [3, 4, 0] with constant -6
• Row 2 (Equation 2): [-1, 0, 3, 1]
• Row 3 (Equation 3): [0, 2, 3, -1]

Thus, the augmented matrix becomes:\[\begin{bmatrix} 3 & 4 & 0 & | & -6 \ -1 & 0 & 3 & | & 1 \ 0 & 2 & 3 & | & -1 \end{bmatrix} \]

Row-Echelon Form

Once we have the augmented matrix, the next goal is to transform it into row-echelon form. Row-echelon form is a triangular form of a matrix that makes it easier to solve for the variables using back substitution. To do this, it is essential to perform row operations:

• Swap rows, if necessary, to place zeros below the pivot position (the leftmost non-zero entry in each row).
• Multiply rows by constants to simplify calculation.
• Add or subtract multiples of rows to eliminate variables from lower rows.

In our example, notice how row operations transform the matrix:

• We swap rows to get a pivot of 1 initially, making the calculations easier.
• We focus on making the entries below each pivot 0, working our way down the matrix.

After performing these operations, the matrix reaches a form where the below-diagonal elements are zeros:\[\begin{bmatrix} -1 & 0 & 3 & | & 1 \ 0 & 4 & 9 & | & -3 \ 0 & 0 & -3 & | & 1 \end{bmatrix}\]This triangular form is crucial for the next step, back substitution.

Back Substitution

After obtaining the row-echelon form of the matrix, the next step in Gaussian Elimination is known as back substitution. This technique involves working from the bottom of the matrix upwards to solve for each variable.

With the triangular form like:\[\begin{cases}0x + 0y - 3z = 1 0x + 4y + 9z = -3 -x + 0y + 3z = 1\end{cases}\]You can start solving for the variables from the last equation first:

• For example, start with the equation for \(z\), which is easy because it only involves one variable: divide through by the coefficient to isolate \(z\).
• Then, substitute \(z\)'s value into the second equation to find \(y\).
• Finally, use both \(y\) and \(z\) in the first equation to determine \(x\).

This back-and-forth substitution continues until all unknowns are isolated, resulting in the solution \(x = -2, y = 0, z = -\frac{1}{3}\). This method ensures that each step builds upon the previous, systematically revealing each variable.

System of Linear Equations

A system of linear equations consists of multiple equations, each representing a straight line in n-dimensional space. Solving such a system means finding a point or set of points that satisfies all equations simultaneously.

In our three-equation system:

• Each equation provides a flat plane in 3D space.
• The solution is the point where these planes intersect.
• Graphically, this intersection point represents the common solution to all equations.

These systems can vary:

• Having a unique solution means the planes intersect at one single point.
• Infinite solutions occur when two or more planes overlap entirely along a line.
• No solutions are present if the planes never intersect.

Understanding the nature of solutions is fundamental to solving linear systems, whether using graphing, substitution, or matrix methods such as Gaussian Elimination. By traversing through the Gaussian Elimination steps, you can systematically find the solution, or determine if one does not exist, in a logical, step-by-step manner.