Math Problem Statement

Let A be a 3x3 matrix given by A = [[-5, -25, 20], [1, 2, -7], [3, 14, -13]]. We want to determine if the columns of matrix A are linearly independent. To do this, we perform row reduction on A as described.

Solution

To determine whether the columns of matrix AA are linearly dependent or independent, we need to perform Gaussian elimination (row reduction). Let's break this down step by step:

The given matrix is:

-5 & -25 & 20 \\ 1 & 2 & -7 \\ 3 & 14 & -13 \end{pmatrix}$$ ### Step 1: Eliminate the first entry in the second and third rows. We want to make the elements below the pivot in the first column zero. - To eliminate the $$1$$ in the second row, we add $$\frac{1}{-5} = -\frac{1}{5}$$ times the first row to the second row. New second row: $$R_2 = R_2 + \left( -\frac{1}{5} \right) R_1$$ Calculation: $$\begin{aligned} 1 + \left( -\frac{1}{5} \right)(-5) &= 0 \\ 2 + \left( -\frac{1}{5} \right)(-25) &= 7 \\ -7 + \left( -\frac{1}{5} \right)(20) &= -11 \end{aligned}$$ So the new second row is $$\begin{pmatrix} 0 & 7 & -11 \end{pmatrix}$$. - To eliminate the $$3$$ in the third row, we add $$\frac{3}{-5} = -\frac{3}{5}$$ times the first row to the third row. New third row: $$R_3 = R_3 + \left( -\frac{3}{5} \right) R_1$$ Calculation: $$\begin{aligned} 3 + \left( -\frac{3}{5} \right)(-5) &= 0 \\ 14 + \left( -\frac{3}{5} \right)(-25) &= -1 \\ -13 + \left( -\frac{3}{5} \right)(20) &= -25 \end{aligned}$$ So the new third row is $$\begin{pmatrix} 0 & -1 & -25 \end{pmatrix}$$. At this point, the matrix is: $$A = \begin{pmatrix} -5 & -25 & 20 \\ 0 & 7 & -11 \\ 0 & -1 & -25 \end{pmatrix}$$ ### Step 2: Eliminate the second entry in the third row. Now, to eliminate the $$-1$$ in the third row, we add $$\frac{-1}{7} = \frac{1}{7}$$ times the second row to the third row. New third row: $$R_3 = R_3 + \left( \frac{1}{7} \right) R_2$$ Calculation: $$\begin{aligned} -1 + \left( \frac{1}{7} \right)(7) &= 0 \\ -25 + \left( \frac{1}{7} \right)(-11) &= -25 - \frac{11}{7} \end{aligned}$$ The third row becomes $$\begin{pmatrix} 0 & 0 & -\frac{186}{7} \end{pmatrix}$$. ### Step 3: Conclusion The matrix is now in row echelon form: $$\begin{pmatrix} -5 & -25 & 20 \\ 0 & 7 & -11 \\ 0 & 0 & -\frac{186}{7} \end{pmatrix}$$ Since the matrix has a pivot in every column, the rank of the matrix is 3, which means that the columns are linearly dependent. Thus, the answer is: - **A. The columns of $$A$$ are linearly dependent.** --- Would you like any further details or clarifications? Here are some related questions for deeper understanding: 1. What is the significance of a pivot column in Gaussian elimination? 2. How do we define linear dependence and independence in the context of column vectors? 3. How would we determine the rank of a matrix if it has dependent columns? 4. Can a square matrix have linearly independent columns but a determinant of zero? 5. How does Gaussian elimination differ from Gauss-Jordan elimination? **Tip:** In row reduction, always focus on eliminating entries systematically to get a triangular matrix, which simplifies determining dependence.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Operations
Gaussian Elimination
Linear Independence

Formulas

Row operations in Gaussian elimination

Theorems

Linear dependence and independence

Suitable Grade Level

Undergraduate Level (College Mathematics)