Math Problem Statement
Solve the following system of equations using reduced row echelon method: 7x + 4y − 5z = 33 5x − 7y − 2z = 37 3x + 7y + 7z = −33 Round to three decimal places if necessary. x = y = z =
Solution
We will solve the given system of linear equations using the reduced row echelon form (RREF) method. The system is:
Step 1: Write the augmented matrix
We can represent this system as an augmented matrix:
7 & 4 & -5 & | & 33 \\ 5 & -7 & -2 & | & 37 \\ 3 & 7 & 7 & | & -33 \end{pmatrix}$$ ### Step 2: Row operations to get the matrix into reduced row echelon form (RREF) #### First, we'll perform the row operations: 1. **Make the first element of the first row (leading 1 in the first column)**: - Row 1 → Row 1 / 7: $$\begin{pmatrix} 1 & \frac{4}{7} & -\frac{5}{7} & | & \frac{33}{7} \\ 5 & -7 & -2 & | & 37 \\ 3 & 7 & 7 & | & -33 \end{pmatrix}$$ 2. **Eliminate the first column elements below the pivot (Row 1)**: - Row 2 → Row 2 - 5 × Row 1 - Row 3 → Row 3 - 3 × Row 1 $$\begin{pmatrix} 1 & \frac{4}{7} & -\frac{5}{7} & | & \frac{33}{7} \\ 0 & -\frac{59}{7} & \frac{23}{7} & | & \frac{116}{7} \\ 0 & \frac{31}{7} & \frac{44}{7} & | & -\frac{240}{7} \end{pmatrix}$$ 3. **Make the pivot in the second row a 1** (divide Row 2 by $$-\frac{59}{7}$$): - Row 2 → Row 2 × $$\frac{-7}{59}$$ $$\begin{pmatrix} 1 & \frac{4}{7} & -\frac{5}{7} & | & \frac{33}{7} \\ 0 & 1 & -\frac{23}{59} & | & \frac{116}{59} \\ 0 & \frac{31}{7} & \frac{44}{7} & | & -\frac{240}{7} \end{pmatrix}$$ 4. **Eliminate the second column elements below and above the pivot (Row 2)**: - Row 1 → Row 1 - $$\frac{4}{7} \times$$ Row 2 - Row 3 → Row 3 - $$\frac{31}{7} \times$$ Row 2 $$\begin{pmatrix} 1 & 0 & -\frac{59}{59} & | & \frac{33}{59} \\ 0 & 1 & -\frac{23}{59} & | & \frac{116}{59} \\ 0 & 0 & \frac{2505}{59} & | & -\frac{18720}{59} \end{pmatrix}$$ 5. **Make the third column pivot a 1** (divide Row 3 by $$\frac{2505}{59}$$): - Row 3 → Row 3 × $$\frac{59}{2505}$$ $$\begin{pmatrix} 1 & 0 & 0 & | & \frac{33}{59} \\ 0 & 1 & 0 & | & \frac{116}{59} \\ 0 & 0 & 1 & | & -\frac{18720}{2505} \end{pmatrix}$$ 6. **Perform the final row elimination**: - Row 1 → Row 1 + Row 3 - Row 2 → Row 2 + Row 3 The final RREF form is: $$\begin{pmatrix} 1 & 0 & 0 & | & \text{2.467} \\ 0 & 1 & 0 & | & \text{1.968} \\ 0 & 0 & 1 & | & -7.472 \end{pmatrix}$$ ### Step 3: Extract the solutions From the final matrix, we can read the values of $$x$$, $$y$$, and $$z$$: $$x = 2.467, \quad y = 1.968, \quad z = -7.472$$ Thus, the solution to the system is: $$x = 2.467, \quad y = 1.968, \quad z = -7.472$$ Would you like more details or have any questions about the solution? Here are some related questions to help further your understanding: 1. What is the significance of the reduced row echelon form (RREF) in solving systems of equations? 2. How can you check if a solution is correct by substituting the values back into the original system of equations? 3. How do you perform Gaussian elimination, and how does it compare to the RREF method? 4. What would happen if the system of equations were inconsistent or had infinitely many solutions? 5. Can this method be applied to systems with more than three variables? How? **Tip:** Double-check your row operations during matrix manipulation to avoid simple errors.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Equations
Row Echelon Form
Gaussian Elimination
Formulas
Augmented Matrix Representation
Row Operations in Matrix Form
Theorems
Reduced Row Echelon Form (RREF)
Gaussian Elimination
Suitable Grade Level
Grades 10-12