Math Problem Statement
Reduce the matrix to its row-reduced form.
9 3 −6 −36
6 8 6 −52 0 2 6 −28
Solution
Let's reduce the given matrix to its row-reduced echelon form (RREF). The matrix is:
9 & 3 & -6 & -36 \\ 6 & 8 & 6 & -52 \\ 0 & 2 & 6 & -28 \end{bmatrix}$$ ### Step 1: Form the augmented matrix: $$\left[ \begin{array}{ccc|c} 9 & 3 & -6 & -36 \\ 6 & 8 & 6 & -52 \\ 0 & 2 & 6 & -28 \end{array} \right]$$ ### Step 2: Make the first pivot entry a 1 by dividing the first row by 9: $$R_1 \to \frac{R_1}{9}$$ $$\left[ \begin{array}{ccc|c} 1 & \frac{1}{3} & -\frac{2}{3} & -4 \\ 6 & 8 & 6 & -52 \\ 0 & 2 & 6 & -28 \end{array} \right]$$ ### Step 3: Eliminate the first entry of the second row using row operations: $$R_2 \to R_2 - 6R_1$$ $$\left[ \begin{array}{ccc|c} 1 & \frac{1}{3} & -\frac{2}{3} & -4 \\ 0 & 6 & 10 & -28 \\ 0 & 2 & 6 & -28 \end{array} \right]$$ ### Step 4: Make the second pivot a 1 by dividing the second row by 6: $$R_2 \to \frac{R_2}{6}$$ $$\left[ \begin{array}{ccc|c} 1 & \frac{1}{3} & -\frac{2}{3} & -4 \\ 0 & 1 & \frac{5}{3} & -\frac{14}{3} \\ 0 & 2 & 6 & -28 \end{array} \right]$$ ### Step 5: Eliminate the second entry of the third row using row operations: $$R_3 \to R_3 - 2R_2$$ $$\left[ \begin{array}{ccc|c} 1 & \frac{1}{3} & -\frac{2}{3} & -4 \\ 0 & 1 & \frac{5}{3} & -\frac{14}{3} \\ 0 & 0 & \frac{4}{3} & -\frac{10}{3} \end{array} \right]$$ ### Step 6: Make the third pivot a 1 by multiplying the third row by $$\frac{3}{4}$$: $$R_3 \to \frac{3}{4}R_3$$ $$\left[ \begin{array}{ccc|c} 1 & \frac{1}{3} & -\frac{2}{3} & -4 \\ 0 & 1 & \frac{5}{3} & -\frac{14}{3} \\ 0 & 0 & 1 & -\frac{5}{2} \end{array} \right]$$ ### Step 7: Eliminate the third entry in the second row and the third entry in the first row: $$R_2 \to R_2 - \frac{5}{3}R_3$$ $$\left[ \begin{array}{ccc|c} 1 & \frac{1}{3} & -\frac{2}{3} & -4 \\ 0 & 1 & 0 & \frac{4}{3} \\ 0 & 0 & 1 & -\frac{5}{2} \end{array} \right]$$ $$R_1 \to R_1 + \frac{2}{3}R_3$$ $$\left[ \begin{array}{ccc|c} 1 & \frac{1}{3} & 0 & -\frac{10}{3} \\ 0 & 1 & 0 & \frac{4}{3} \\ 0 & 0 & 1 & -\frac{5}{2} \end{array} \right]$$ ### Step 8: Eliminate the second entry of the first row: $$R_1 \to R_1 - \frac{1}{3}R_2$$ $$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & \frac{4}{3} \\ 0 & 0 & 1 & -\frac{5}{2} \end{array} \right]$$ This is the row-reduced echelon form (RREF) of the matrix. The solution to the system of equations represented by this augmented matrix is: $$x_1 = -2, \quad x_2 = \frac{4}{3}, \quad x_3 = -\frac{5}{2}$$ Would you like any further clarifications? ### Related Questions: 1. How do you perform row-reduction for a larger matrix? 2. What is the significance of row-reduced echelon form (RREF)? 3. How can row-reduction help solve systems of equations? 4. What happens if a matrix has no row-reduced form? 5. Can row-reduction be used for non-square matrices? **Tip:** Always ensure the pivot elements (leading 1s) are in the correct columns to reach RREF successfully!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Row-Reduction
Matrix Operations
Systems of Equations
Formulas
Row-reduced echelon form (RREF) transformation rules
Theorems
Gauss-Jordan Elimination
Suitable Grade Level
Grades 10-12
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