Math Problem Statement

Use row operations to change the matrix to reduced form. left bracket Start 1 By 2 Table 1st Row 1st Column Start 2 By 3 Matrix 1st Row 1st Column 1 2nd Column 1 3rd Column 1 2nd Row 1st Column 2 2nd Column 3 3rd Column 6 EndMatrix 2nd Column Start 2 By 1 Matrix 1st Row 1st Column 17 2nd Row 1st Column 27 EndMatrix EndTable right bracket 1 1 1 2 3 6 17 27 Question content area bottom Part 1 left bracket Start 1 By 2 Table 1st Row 1st Column Start 2 By 3 Matrix 1st Row 1st Column 1 2nd Column 1 3rd Column 1 2nd Row 1st Column 2 2nd Column 3 3rd Column 6 EndMatrix 2nd Column Start 2 By 1 Matrix 1st Row 1st Column 17 2nd Row 1st Column 27 EndMatrix EndTable right bracket 1 1 1 2 3 6 17 27 ​~ left bracket Start 2 By 4 Table 1st Row 1st Column nothing 2nd Column nothing 3rd Column nothing 4st Column nothing 2nd Row 1st Column nothing 2nd Column nothing 3rd Column nothing 4st Column nothing EndTable right bracket

Solution

Let's solve this step by step using row operations to reduce the matrix to its reduced row echelon form.

Step 1: Write the augmented matrix

We are given the matrix:

1 & 1 & 1 & | & 17 \\ 2 & 3 & 6 & | & 27 \end{bmatrix}$$ This is a 2x3 matrix with an augmented column for the constants (17, 27). ### Step 2: Apply row operations We will use elementary row operations to reduce this matrix to its **reduced row echelon form**. #### Operation 1: Make the first pivot element a 1 In this case, the first element in the first row is already 1, so no changes are needed for the first row. $$\begin{bmatrix} 1 & 1 & 1 & | & 17 \\ 2 & 3 & 6 & | & 27 \end{bmatrix}$$ #### Operation 2: Eliminate the 2 in the second row, first column We need to eliminate the 2 in the second row, first column, by subtracting $$2R_1$$ from $$R_2$$ (i.e., the second row). This gives us the new second row: \[ R_2 = R_2 - 2R_1 = [2 - 2(1), 3 - 2(1), 6 - 2(1), 27 - 2(17)] = [0, 1, 4, -7] \] So the matrix now becomes: $$\begin{bmatrix} 1 & 1 & 1 & | & 17 \\ 0 & 1 & 4 & | & -7 \end{bmatrix}$$ #### Operation 3: Eliminate the 1 in the first row, second column To eliminate the 1 in the first row, second column, we will subtract $$R_2$$ from $$R_1$$: \[ R_1 = R_1 - R_2 = [1 - 0, 1 - 1, 1 - 4, 17 - (-7)] = [1, 0, -3, 24] \] Thus, the new matrix is: $$\begin{bmatrix} 1 & 0 & -3 & | & 24 \\ 0 & 1 & 4 & | & -7 \end{bmatrix}$$ #### Operation 4: Make the third column of the first row a 1 To simplify further, we divide the first row by $$-3$$ to make the third element in the first row a 1: \[ R_1 = \frac{1}{-3} R_1 = \left[ \frac{1}{-3} \times 1, \frac{1}{-3} \times 0, \frac{1}{-3} \times -3, \frac{1}{-3} \times 24 \right] = \left[ -\frac{1}{3}, 0, 1, -8 \right] \] Thus, the final matrix is: $$\begin{bmatrix} 1 & 0 & 1 & | & -8 \\ 0 & 1 & 4 & | & -7 \end{bmatrix}$$ This is the **reduced row echelon form** of the matrix. ### Final Matrix: $$\begin{bmatrix} 1 & 0 & 1 & | & -8 \\ 0 & 1 & 4 & | & -7 \end{bmatrix}$$ Would you like further details or explanations? ### 5 Related Questions: 1. What are the key properties of a matrix in reduced row echelon form? 2. How do you interpret the solution of the system from this reduced matrix? 3. Can you use Gaussian elimination for larger matrices in the same way? 4. What are the types of elementary row operations, and how do they affect the determinant of a matrix? 5. How can you verify if a system of equations has infinite solutions or no solution from its matrix? ### Tip: Always aim to identify and work with pivot elements (leading 1's) early on when performing row reductions to streamline the process.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Operations
Row Reduction
Gaussian Elimination

Formulas

Elementary row operations
Reduced Row Echelon Form

Theorems

Gaussian Elimination
Row Equivalence Theorem

Suitable Grade Level

Grades 11-12, College Introductory

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