Math Problem Statement
giải
Consider the case where m = 1, and in the vector space R4, the subspace V is the set of solutions to the system AX = 0. Find a basis for V.
Solution
We are given a matrix and asked to find a basis for the null space , which consists of the solutions to the equation , where and . Let me extract the matrix and solve the system step by step.
Matrix is:
1 & 2 & -2 & -1 \\ 2 & 1 & 1 & 3 \\ 1 & 5 & -7 & -6 \\ -3 & 0 & -4 & 1 \end{bmatrix}$$ We need to find the null space of this matrix, which is the set of all vectors $$X \in \mathbb{R}^4$$ such that $$AX = 0$$. ### Step 1: Solve $$AX = 0$$ The system of equations from $$AX = 0$$ is equivalent to solving this matrix equation. We will perform Gaussian elimination (or row reduction) to bring the matrix $$A$$ to its reduced row echelon form (RREF) and then find the solution. Let's proceed with the calculation: 1. Subtract 2 times row 1 from row 2, subtract 1 times row 1 from row 3, and add 3 times row 1 to row 4. Resulting matrix: $$\begin{bmatrix} 1 & 2 & -2 & -1 \\ 0 & -3 & 5 & 5 \\ 0 & 3 & -5 & -5 \\ 0 & 6 & -10 & -2 \end{bmatrix}$$ 2. Now, subtract row 2 from row 3 and subtract 2 times row 2 from row 4. Resulting matrix: $$\begin{bmatrix} 1 & 2 & -2 & -1 \\ 0 & -3 & 5 & 5 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -12 \end{bmatrix}$$ 3. Divide row 2 by -3 and row 4 by -12. Final matrix in RREF: $$\begin{bmatrix} 1 & 2 & -2 & -1 \\ 0 & 1 & -\frac{5}{3} & -\frac{5}{3} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ ### Step 2: Interpretation From the RREF, we can derive the following system of equations: $$\begin{aligned} x_1 + 2x_2 - 2x_3 - x_4 &= 0 \\ x_2 - \frac{5}{3}x_3 - \frac{5}{3}x_4 &= 0 \\ x_4 &= 0 \end{aligned}$$ Since $$x_4 = 0$$, substitute it into the other equations to simplify: $$\begin{aligned} x_1 + 2x_2 - 2x_3 &= 0 \\ x_2 - \frac{5}{3}x_3 &= 0 \end{aligned}$$ From $$x_2 = \frac{5}{3}x_3$$, substitute this into the first equation: $$x_1 + 2\left(\frac{5}{3}x_3\right) - 2x_3 = 0 \implies x_1 + \frac{10}{3}x_3 - 2x_3 = 0 \implies x_1 = -\frac{4}{3}x_3$$ Thus, the general solution for the system is: $$x_1 = -\frac{4}{3}x_3, \quad x_2 = \frac{5}{3}x_3, \quad x_4 = 0$$ Therefore, the solution vector is: $$X = x_3 \begin{bmatrix} -\frac{4}{3} \\ \frac{5}{3} \\ 1 \\ 0 \end{bmatrix}$$ ### Step 3: Find a basis for $$V$$ The null space is spanned by the vector: $$\left\{ \begin{bmatrix} -4 \\ 5 \\ 3 \\ 0 \end{bmatrix} \right\}$$ Thus, a basis for $$V$$ is $$\left\{ (-4, 5, 3, 0) \right\}$$, corresponding to option **B** in the multiple-choice answers. ### Conclusion The correct answer is **B: $$\left\{ (-4, 5, 3, 0) \right\}$$**. Would you like more details on any step, or do you have any further questions? Here are five related questions to deepen your understanding: 1. How is the null space of a matrix defined? 2. Why do we use Gaussian elimination to solve systems of linear equations? 3. What are the steps involved in row reduction to get a matrix into RREF? 4. How can we verify if a vector is in the null space of a matrix? 5. What does it mean if the null space of a matrix is trivial? **Tip:** Always double-check row operations during Gaussian elimination to avoid calculation errors.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Null Space
Gaussian Elimination
Matrix Operations
Formulas
AX = 0 (homogeneous system)
Gaussian elimination to bring matrix to RREF
Theorems
The Null Space Theorem
Row Reduction
Suitable Grade Level
Undergraduate Linear Algebra (Grades 11-12 and above)