Math Problem Statement
Using the techniques discussed in this section, solve the following system of linear equations. Enter the solution as an ordered triple. For parametric solutions use
z = t
as the parameter. If a solution does not exist, enter DNE for all coordinates.
−3x + 7y − 8z
=
9
x − y + 2z
=
−9
−4y + z
=
7
Solution
To solve the given system of linear equations, we'll use a combination of elimination and substitution methods. We want to find values for , , and that satisfy all three equations. Here are the equations provided:
-3x + 7y - 8z = 9 \\ x - y + 2z = -9 \\ -4y + z = 7 \end{cases}$$ ### Step 1: Write the System in Matrix Form We can express the system as an augmented matrix: $$\left[ \begin{array}{ccc|c} -3 & 7 & -8 & 9 \\ 1 & -1 & 2 & -9 \\ 0 & -4 & 1 & 7 \end{array} \right]$$ ### Step 2: Use Gaussian Elimination to Find the Solution We'll perform row operations to bring the matrix to reduced row echelon form (RREF). Let's proceed with the steps: 1. **Swap Row 1 and Row 2** to have a leading 1 in the first row: $$\left[ \begin{array}{ccc|c} 1 & -1 & 2 & -9 \\ -3 & 7 & -8 & 9 \\ 0 & -4 & 1 & 7 \end{array} \right]$$ 2. **Add 3 times Row 1 to Row 2**: $$\left[ \begin{array}{ccc|c} 1 & -1 & 2 & -9 \\ 0 & 4 & -2 & -18 \\ 0 & -4 & 1 & 7 \end{array} \right]$$ 3. **Add Row 2 to Row 3**: $$\left[ \begin{array}{ccc|c} 1 & -1 & 2 & -9 \\ 0 & 4 & -2 & -18 \\ 0 & 0 & -1 & -11 \end{array} \right]$$ 4. **Multiply Row 3 by -1** to get a leading 1: $$\left[ \begin{array}{ccc|c} 1 & -1 & 2 & -9 \\ 0 & 4 & -2 & -18 \\ 0 & 0 & 1 & 11 \end{array} \right]$$ 5. **Add 2 times Row 3 to Row 2**: $$\left[ \begin{array}{ccc|c} 1 & -1 & 2 & -9 \\ 0 & 4 & 0 & 4 \\ 0 & 0 & 1 & 11 \end{array} \right]$$ 6. **Divide Row 2 by 4**: $$\left[ \begin{array}{ccc|c} 1 & -1 & 2 & -9 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 11 \end{array} \right]$$ 7. **Add Row 2 to Row 1**: $$\left[ \begin{array}{ccc|c} 1 & 0 & 2 & -8 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 11 \end{array} \right]$$ 8. **Subtract 2 times Row 3 from Row 1**: $$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & -30 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 11 \end{array} \right]$$ ### Step 3: Interpret the Results The matrix is now in reduced row echelon form, corresponding to the following system of equations: $$\begin{cases} x = -30 \\ y = 1 \\ z = 11 \end{cases}$$ ### Solution The solution as an ordered triple is: $$(x, y, z) = (-30, 1, 11)$$ ### Conclusion The solution to the system of equations is $$(-30, 1, 11)$$. If you have any questions or want further details about any step, feel free to ask! #### Related Questions: 1. How can we verify the solution of a system of linear equations? 2. What are other methods to solve a system of linear equations apart from Gaussian elimination? 3. How does the augmented matrix method help in finding solutions to linear systems? 4. What is the difference between Row Echelon Form (REF) and Reduced Row Echelon Form (RREF)? 5. What are the possible outcomes when solving a system of linear equations? #### Tip: Always double-check each row operation when using Gaussian elimination to ensure no arithmetic mistakes are made.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Formulas
Matrix form of linear equations
Gaussian elimination to reduce to Row Echelon Form (REF)
Reduced Row Echelon Form (RREF)
Theorems
Gaussian Elimination Method
Row Echelon Form (REF)
Reduced Row Echelon Form (RREF)
Suitable Grade Level
Grades 10-12
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