Math Problem Statement
calculate correctly
Find the solution set of the given linear system and explain why the set of vectors does or does not span R^4.
Solution
Let's solve the given problems step-by-step.
Given Matrix
The transformation is defined by the matrix :
1 & -3 & 1 & 2 & 2 \\ -1 & 3 & 0 & -2 & -2 \\ 1 & -3 & 3 & 3 & 4 \\ 0 & 0 & 5 & -4 & -8 \end{bmatrix}$$ The Reduced Row Echelon Form (RREF) of $$A$$ is given as: $$\text{RREF}(A) = \begin{bmatrix} 1 & -3 & 0 & 0 & -2 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ ### Part (a): Solution Set of the Given System The system of linear equations is: $$\begin{cases} x_1 - 3x_2 + x_3 + 2x_4 + 2x_5 = 0 \\ -x_1 + 3x_2 + 0x_3 - 2x_4 - 2x_5 = 0 \\ x_1 - 3x_2 + 3x_3 + 3x_4 + 4x_5 = 0 \\ 0x_1 + 0x_2 + 5x_3 - 4x_4 - 8x_5 = 0 \end{cases}$$ We use the RREF form to find the solutions: $$\begin{cases} x_1 - 3x_2 - 2x_5 = 0 \\ x_3 = 0 \\ x_4 + 2x_5 = 0 \end{cases}$$ From the second and third equations, we have: $$x_3 = 0$$ $$x_4 = -2x_5$$ Substitute $$x_3 = 0$$ and $$x_4 = -2x_5$$ into the first equation: $$x_1 - 3x_2 - 2x_5 = 0$$ $$x_1 = 3x_2 + 2x_5$$ Thus, the solution set can be parameterized by $$x_2$$ and $$x_5$$: $$x_1 = 3x_2 + 2x_5$$ $$x_3 = 0$$ $$x_4 = -2x_5$$ Therefore, the general solution is: $$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{pmatrix} = \begin{pmatrix} 3x_2 + 2x_5 \\ x_2 \\ 0 \\ -2x_5 \\ x_5 \end{pmatrix} = x_2 \begin{pmatrix} 3 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + x_5 \begin{pmatrix} 2 \\ 0 \\ 0 \\ -2 \\ 1 \end{pmatrix}$$ ### Part (b): Spanning Set and Basis The set $$S$$ is: $$S = \left\{ \begin{pmatrix} 1 \\ -1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -3 \\ 3 \\ -3 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 3 \\ 5 \end{pmatrix}, \begin{pmatrix} 2 \\ -2 \\ 3 \\ -4 \end{pmatrix} \right\}$$ To determine if $$S$$ spans $$\mathbb{R}^4$$, we need to check the rank of the matrix formed by these vectors as columns: $$B = \begin{bmatrix} 1 & -3 & 1 & 2 \\ -1 & 3 & 0 & -2 \\ 1 & -3 & 3 & 3 \\ 0 & 0 & 5 & -4 \end{bmatrix}$$ We see from the RREF of $$A$$: $$\text{RREF}(A) = \begin{bmatrix} 1 & -3 & 0 & 0 & -2 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ Since the rank of $$A$$ (and hence $$B$$) is 3 (there are 3 non-zero rows), the set $$S$$ does not span $$\mathbb{R}^4$$ because $$\mathbb{R}^4$$ requires 4 linearly independent vectors. Would you like further details or have any questions? Here are some related questions for practice: 1. How do you determine the rank of a matrix? 2. What is the significance of the RREF of a matrix in solving linear systems? 3. How can you determine if a set of vectors spans a given space? 4. What does it mean for vectors to be linearly independent? 5. How do you find a basis for the null space of a matrix? **Tip:** To check if a set of vectors spans a space, always consider the number of vectors and their linear independence.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Linear Equations
Vector Spaces
Matrix Rank
Formulas
RREF (Reduced Row Echelon Form)
General Solution to Linear Systems
Conditions for Spanning a Vector Space
Theorems
Rank-Nullity Theorem
Linear Independence and Dependence
Basis of a Vector Space
Suitable Grade Level
Undergraduate
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