Math Problem Statement
Solve using row reduction method. 5x-6y-7z=7, 6x-4y-10z=-34, 2x+4y-3z=29. Show detailed solution
Solution
We will solve the system of equations using row reduction (Gaussian elimination) to obtain the solution for , , and .
Step 1: Convert the System into an Augmented Matrix
The given system of equations is:
5x - 6y - 7z &= 7 \\ 6x - 4y - 10z &= -34 \\ 2x + 4y - 3z &= 29 \end{aligned}$$ The augmented matrix is: $$\begin{bmatrix} 5 & -6 & -7 & | 7 \\ 6 & -4 & -10 & | -34 \\ 2 & 4 & -3 & | 29 \end{bmatrix}$$ ### **Step 2: Make the First Pivot Element 1** The first pivot is already **5** in position $$(1,1)$$, but we prefer it to be **1**. We divide **Row 1** by **5**: $$\begin{bmatrix} 1 & -\frac{6}{5} & -\frac{7}{5} & | \frac{7}{5} \\ 6 & -4 & -10 & | -34 \\ 2 & 4 & -3 & | 29 \end{bmatrix}$$ ### **Step 3: Eliminate First Column Below the Pivot** To eliminate the first column below the pivot, we use row operations: - **Row 2 → Row 2 - 6 × Row 1** - **Row 3 → Row 3 - 2 × Row 1** $$\begin{bmatrix} 1 & -\frac{6}{5} & -\frac{7}{5} & | \frac{7}{5} \\ 0 & \frac{26}{5} & -\frac{8}{5} & | -\frac{112}{5} \\ 0 & \frac{34}{5} & -\frac{1}{5} & | \frac{123}{5} \end{bmatrix}$$ ### **Step 4: Make the Second Pivot Element 1** We make the pivot **1** by multiplying Row 2 by **5/26**: $$\begin{bmatrix} 1 & -\frac{6}{5} & -\frac{7}{5} & | \frac{7}{5} \\ 0 & 1 & -\frac{4}{13} & | -\frac{112}{26} \\ 0 & \frac{34}{5} & -\frac{1}{5} & | \frac{123}{5} \end{bmatrix}$$ ### **Step 5: Eliminate Second Column Below and Above the Pivot** We update **Row 1** and **Row 3**: - **Row 1 → Row 1 + (6/5) × Row 2** - **Row 3 → Row 3 - (34/5) × Row 2** $$\begin{bmatrix} 1 & 0 & -\frac{5}{2} & | \frac{9}{2} \\ 0 & 1 & -\frac{4}{13} & | -\frac{56}{13} \\ 0 & 0 & -\frac{25}{13} & | \frac{985}{13} \end{bmatrix}$$ ### **Step 6: Make the Third Pivot Element 1** We multiply **Row 3** by **-13/25** to get 1: $$\begin{bmatrix} 1 & 0 & -\frac{5}{2} & | \frac{9}{2} \\ 0 & 1 & -\frac{4}{13} & | -\frac{56}{13} \\ 0 & 0 & 1 & | -39 \end{bmatrix}$$ ### **Step 7: Back Substitution** We solve for **$$z$$, $$y$$, and $$x$$**: - **From Row 3:** $$z = -39$$ - **From Row 2:** $$y - \frac{4}{13}(-39) = -\frac{56}{13}$$ $$y + \frac{156}{13} = -\frac{56}{13}$$ $$y = -\frac{212}{13} = -16$$ - **From Row 1:** $$x - \frac{5}{2}(-39) = \frac{9}{2}$$ $$x + \frac{195}{2} = \frac{9}{2}$$ $$x = -\frac{186}{2} = -93$$ ### **Final Answer:** $$\boxed{(x, y, z) = (-93, -16, -39)}$$ Would you like further clarification on any step? Here are five related problems for practice: 1. Solve using row reduction: $$\begin{aligned} 2x + y - 3z &= 5 \\ x - 4y + 2z &= -3 \\ 3x + 2y - z &= 4 \end{aligned}$$ 2. Convert a given system into an augmented matrix. 3. Solve using row echelon form: $$\begin{aligned} x + y + z &= 6 \\ 2x + 3y + z &= 14 \\ 3x + y + 2z &= 14 \end{aligned}$$ 4. Explain the difference between row echelon and reduced row echelon form. 5. Solve a 4×4 system using Gaussian elimination. **Tip:** Always check your row operations carefully to avoid arithmetic mistakes!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Gaussian Elimination
Row Reduction
Systems of Equations
Formulas
Gaussian Elimination
Row Operations: Row2 → Row2 - k × Row1
Back Substitution
Theorems
Row Reduction Method (Gaussian Elimination)
Pivoting in Matrix Operations
Suitable Grade Level
Grades 9-12