Math Problem Statement

Let A = [[2, 0, 6], [-1, 8, 5], [1, -2, 1]] and b = [10, 3, 3], and let W be the set of all linear combinations of the columns of A. Determine if b is in W.

Solution

Problem Analysis:

We are given the matrix A=[206185121]A = \begin{bmatrix} 2 & 0 & 6 \\ -1 & 8 & 5 \\ 1 & -2 & 1 \end{bmatrix}, and the vector b=[1033]\mathbf{b} = \begin{bmatrix} 10 \\ 3 \\ 3 \end{bmatrix}.
The set WW is defined as the span of the columns of AA, i.e., the set of all linear combinations of the columns of AA. The goal is to determine if b\mathbf{b} is in WW.

This is equivalent to solving the matrix equation: Ax=b,A\mathbf{x} = \mathbf{b}, where x=[x1x2x3]\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} are the coefficients of the linear combination of the columns of AA. If a solution exists, then b\mathbf{b} is in WW; otherwise, it is not.


Solving the System:

We solve Ax=bA\mathbf{x} = \mathbf{b}:

\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 10 \\ 3 \\ 3 \end{bmatrix}.$$ This can be solved by writing the augmented matrix and performing row reduction. $$\text{Augmented matrix:} \quad \left[\begin{array}{ccc|c} 2 & 0 & 6 & 10 \\ -1 & 8 & 5 & 3 \\ 1 & -2 & 1 & 3 \end{array}\right].$$ --- Let me compute the solution step-by-step. numpy as np # Define matrix A and vector b A = np.array([[2, 0, 6], [-1, 8, 5], [1, -2, 1]]) b = np.array([10, 3, 3]) # Solve the augmented matrix using numpy.linalg.lstsq to determine if a solution exists solution, residuals, rank, singular_values = np.linalg.lstsq(A, b, rcond=None) # Check if the solution satisfies the equation A*x = b (or if b is in the column space of A) b_is_in_span = np.allclose(np.dot(A, solution), b) solution, residuals, rank, singular_values, b_is_in_span### Solution: The system $$A\mathbf{x} = \mathbf{b}$$ has a solution. The computed solution is: $$\mathbf{x} = \begin{bmatrix} 0.6364 \\ -0.4545 \\ 1.4545 \end{bmatrix}.$$ The rank of $$A$$ is 2, and the vector $$\mathbf{b}$$ lies within the column space (span of the columns) of $$A$$. Therefore, $$\mathbf{b} \in W$$. --- ### Conclusion: Since $$\mathbf{b}$$ is in the span of the columns of $$A$$, we conclude that: $$\mathbf{b} \in W.$$ Would you like a step-by-step explanation of the row reduction, or more details on the solution process? Let me know! --- ### Related Questions: 1. What does it mean geometrically for $$\mathbf{b}$$ to be in the span of the columns of $$A$$? 2. How would the solution change if the rank of $$A$$ was less than 3? 3. How is the rank of $$A$$ related to the existence of solutions? 4. What happens if $$\mathbf{b}$$ is not in $$W$$? Can we still compute a "least-squares" solution? 5. How can we verify computationally that $$\mathbf{b}$$ is in $$W$$? ### Tip: When solving systems of linear equations, checking the rank of the matrix helps you understand whether the system is consistent (has a solution) or inconsistent (no solution).

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Equations
Span of Vectors

Formulas

A * x = b
Row Reduction of Augmented Matrices

Theorems

Span of a Set of Vectors
Rank-Nullity Theorem

Suitable Grade Level

College Level (Undergraduate)