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Chapter 9: Problem 30

Solve the system of equations using Gaussian elimination or Gauss-Jordan elimination. Use a graphing calculator to check your answer. $$\begin{aligned} 3 x+2 y+2 z &=3 \\ x+2 y-z &=5 \\ 2 x-4 y+z &=0 \end{aligned}$$

Short Answer

Expert verified
The solution is \(x = \frac{7}{2}\), \(y = -\frac{1}{4}\), and \(z = -2\).

Step by step solution

01

- Write the system as an augmented matrix

Convert the system of linear equations into an augmented matrix. The system is: \[ \begin{aligned} 3x + 2y + 2z &= 3 \ x + 2y - z &= 5 \ 2x - 4y + z &= 0 \end{aligned} \] So the augmented matrix is: \[ \begin{bmatrix} 3 & 2 & 2 & | & 3 \ 1 & 2 & -1 & | & 5 \ 2 & -4 & 1 & | & 0 \end{bmatrix} \]
02

- Transform to upper triangular form

Perform row operations to transform the matrix into upper triangular form. First, make the element in the first column, second row a zero by subtracting the first row divided by 3 from the second row:\[ \text{R2} \rightarrow \text{R2} - \frac{1}{3} \text{R1} = \begin{bmatrix} 1 & 2 & -1 & | & 5 \end{bmatrix} - \frac{1}{3} \begin{bmatrix} 3 & 2 & 2 & | & 3 \end{bmatrix} = \begin{bmatrix} 0 & \frac{4}{3} & -\frac{5}{3} & | & \frac{12}{3} \end{bmatrix} \]Now, make the element in the first column, third row a zero by subtracting twice the first row from the third row:\[ \text{R3} \rightarrow \text{R3} - 2 \text{R1} = \begin{bmatrix} 2 & -4 & 1 & | & 0 \end{bmatrix} - 2 \begin{bmatrix} 3 & 2 & 2 & | & 3 \end{bmatrix} = \begin{bmatrix} -4 & -8 & -3 & | & -6 \end{bmatrix} \]
03

- Simplify second and third rows

Make the leading coefficient of the new second row 1 by multiplying it by 3/4. The second row becomes: \[ \begin{bmatrix} 0 & 1 & -\frac{5}{4} & | & \frac{9}{4} \end{bmatrix} \]Make the leading coefficient of the new third row 1 by multiplying it by -1/4.The third row becomes: \[\begin{bmatrix} 0 & 2 & \frac{1}{4} & | & \frac{3}{2} \end{bmatrix}\]
04

- Make the element in the second column of the third row zero

Subtract 2 times the second row from the third row to make the element in the second column of the third row zero: \[ \text{R3} \rightarrow \text{R3} - 2 \times \text{R2} = \begin{bmatrix} 0 & 2 & \frac{1}{4} & | & \frac{3}{2} \end{bmatrix} - 2 \begin{bmatrix} 0 & 1 & -\frac{5}{4} & | & \frac{9}{4} \end{bmatrix} = \begin{bmatrix} 0 & 0 & 3 & | & -6 \end{bmatrix} \]
05

- Perform back substitution

Now, solve the system starting from the third row upward using back substitution. From the third row: \[ 3z = -6 \rightarrow z = -2 \]Substitute \(z = -2\) into the second row: \[ y - \frac{5}{4}(-2) = \frac{9}{4} \rightarrow y + \frac{5}{2} = \frac{9}{4} \rightarrow y = \frac{9}{4} - \frac{10}{4} = -\frac{1}{4} \]Substitute \(y = -\frac{1}{4}\) and \(z = -2\) into the first row: \[ 3x + 2(-\frac{1}{4}) + 2(-2) = 3 \rightarrow 3x - \frac{1}{2} - 4 = 3 \rightarrow 3x = 3 + 4 + \frac{1}{2} = \frac{21}{2} \rightarrow x = \frac{7}{2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

System of Linear Equations
A system of linear equations is a set of equations where each equation is linear, meaning no variable is raised to a power greater than one. Solving these systems involves finding the values of the variables that satisfy all equations simultaneously.

The given system of equations is:
\[ \begin{aligned} 3x + 2y + 2z &= 3 \ x + 2y - z &= 5 \ 2x - 4y + z &= 0 \end{aligned} \]

Solving such systems can be done using different methods, and one effective way is using Gaussian Elimination. This method transforms the system into a form that makes it easier to solve, step by step.
Augmented Matrix
An augmented matrix is used to represent a system of linear equations. It combines the coefficients of the variables and the constants into a single matrix format, making it easier to apply row operations.

The augmented matrix for our system is:
\[ \begin{bmatrix} 3 & 2 & 2 & | & 3 \ 1 & 2 & -1 & | & 5 \ 2 & -4 & 1 & | & 0 \end{bmatrix} \]

Each row corresponds to one equation, with columns for each variable and the constant term. The vertical bar separates the coefficients from the constants. Transforming the augmented matrix involves using row operations to achieve an upper triangular form, making it easier to find solutions.
Back Substitution
Back substitution is the process of solving a system of equations that has been transformed into an upper triangular form. Starting from the last equation, which contains only one variable, you solve for that variable and then substitute its value back into the previous equations to find the remaining variables.

In our solution, we transformed the augmented matrix to an upper triangular form, and the final steps include:
  • From the third row: \[ 3z = -6 \rightarrow z = -2 \]
  • Substituting \( z = -2 \) into the second row: \[ y - \frac{5}{4}(-2) = \frac{9}{4} \rightarrow y = -\frac{1}{4} \]
  • Substituting \( y = -\frac{1}{4} \) and \( z = -2 \) into the first row: \[ 3x + 2(-\frac{1}{4}) + 2(-2) = 3 \rightarrow x = \frac{7}{2} \]

By systematically substituting values back into earlier equations, we obtain the complete solution for the system: \( x = \frac{7}{2} \), \( y = -\frac{1}{4} \), \( z = -2 \).

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Most popular questions from this chapter

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