Math Problem Statement
([[1, 5, 4, 6], [2, 8, 4, 1], [1, 4, 2, 1]])
What is the reduced row echelon form of the following augmented matrix
Solution
We are given the augmented matrix:
1 & 5 & 4 & 6 \\ 2 & 8 & 4 & 1 \\ 1 & 4 & 2 & 1 \end{bmatrix}$$ To reduce this matrix to its reduced row echelon form (RREF), we perform a series of row operations (such as swapping rows, multiplying rows by constants, and adding/subtracting rows) until we reach the desired form. Let's go through the steps: ### Step 1: Make the element in the first row, first column a leading 1 This is already true since $$a_{11} = 1$$. ### Step 2: Eliminate the elements below the pivot in the first column We need to eliminate the elements in the second and third rows, first column (2 and 1). - Subtract $$2 \times \text{Row 1}$$ from Row 2: \[ R2 \to R2 - 2R1 = [2, 8, 4, 1] - 2 \times [1, 5, 4, 6] = [0, -2, -4, -11] \] - Subtract $$1 \times \text{Row 1}$$ from Row 3: \[ R3 \to R3 - R1 = [1, 4, 2, 1] - 1 \times [1, 5, 4, 6] = [0, -1, -2, -5] \] Now the matrix looks like this: $$\begin{bmatrix} 1 & 5 & 4 & 6 \\ 0 & -2 & -4 & -11 \\ 0 & -1 & -2 & -5 \end{bmatrix}$$ ### Step 3: Make the pivot in the second row, second column a leading 1 We need to divide Row 2 by $$-2$$ to make the pivot a leading 1: \[ R2 \to \frac{1}{-2} R2 = [0, -2, -4, -11] \to [0, 1, 2, \frac{11}{2}] \] Now the matrix looks like this: $$\begin{bmatrix} 1 & 5 & 4 & 6 \\ 0 & 1 & 2 & \frac{11}{2} \\ 0 & -1 & -2 & -5 \end{bmatrix}$$ ### Step 4: Eliminate the element below the pivot in the second column We need to eliminate the $$-1$$ in Row 3, second column. To do this, we add Row 2 to Row 3: \[ R3 \to R3 + R2 = [0, -1, -2, -5] + [0, 1, 2, \frac{11}{2}] = [0, 0, 0, \frac{1}{2}] \] Now the matrix looks like this: $$\begin{bmatrix} 1 & 5 & 4 & 6 \\ 0 & 1 & 2 & \frac{11}{2} \\ 0 & 0 & 0 & \frac{1}{2} \end{bmatrix}$$ ### Step 5: Eliminate the element above the pivot in the second column We need to eliminate the $$5$$ in Row 1, second column. To do this, we subtract $$5 \times \text{Row 2}$$ from Row 1: \[ R1 \to R1 - 5R2 = [1, 5, 4, 6] - 5 \times [0, 1, 2, \frac{11}{2}] = [1, 0, -6, -\frac{43}{2}] \] Now the matrix looks like this: $$\begin{bmatrix} 1 & 0 & -6 & -\frac{43}{2} \\ 0 & 1 & 2 & \frac{11}{2} \\ 0 & 0 & 0 & \frac{1}{2} \end{bmatrix}$$ ### Step 6: Scale the third row Finally, we divide Row 3 by $$\frac{1}{2}$$ to make the pivot a 1: \[ R3 \to \frac{1}{\frac{1}{2}} R3 = [0, 0, 0, \frac{1}{2}] \to [0, 0, 0, 1] \] ### Final Reduced Row Echelon Form (RREF): $$\begin{bmatrix} 1 & 0 & -6 & -\frac{43}{2} \\ 0 & 1 & 2 & \frac{11}{2} \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ This is the reduced row echelon form of the given augmented matrix. --- Would you like any additional clarifications or details? Here are some related questions: 1. What are the properties of a matrix in reduced row echelon form? 2. Can a matrix have more than one reduced row echelon form? 3. How is the reduced row echelon form used to solve systems of linear equations? 4. What is the difference between row echelon form and reduced row echelon form? 5. How do elementary row operations affect the determinant of a matrix? **Tip:** In solving linear systems, the RREF method guarantees a unique, simplified form, which makes it easy to identify the solution or check for inconsistencies.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Operations
Row Reduction
Augmented Matrices
Formulas
Elementary Row Operations
Row Echelon Form
Reduced Row Echelon Form (RREF)
Theorems
Gauss-Jordan Elimination
Suitable Grade Level
Grades 10-12 and College
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