Math Problem Statement
(1 point) Let
π΄=β‘β£β’β’β3β20640β6β40β€β¦β₯β₯.A=[β36β6β24β4000].
A basis for the null space of π΄A is
Solution
We are tasked with finding a basis for the null space of the matrix . The matrix is given as:
-3 & 6 & -6 \\ -2 & 4 & -4 \\ 0 & 0 & 0 \end{bmatrix}$$ ### Step 1: Set up the null space equation The null space consists of all vectors $$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$$ such that: $$A \mathbf{x} = \mathbf{0}$$ This gives us the following system of equations: $$\begin{aligned} -3x_1 + 6x_2 - 6x_3 &= 0 \\ -2x_1 + 4x_2 - 4x_3 &= 0 \\ 0x_1 + 0x_2 + 0x_3 &= 0 \end{aligned}$$ ### Step 2: Simplify the system The first two equations are: $$-3x_1 + 6x_2 - 6x_3 = 0 \quad \text{(Equation 1)}$$ $$-2x_1 + 4x_2 - 4x_3 = 0 \quad \text{(Equation 2)}$$ Equation 1 can be simplified by dividing through by $$-3$$: $$x_1 - 2x_2 + 2x_3 = 0 \quad \text{(Simplified Equation 1)}$$ Similarly, Equation 2 can be simplified by dividing through by $$-2$$: $$x_1 - 2x_2 + 2x_3 = 0 \quad \text{(Simplified Equation 2)}$$ Since both simplified equations are identical, we are left with the single equation: $$x_1 = 2x_2 - 2x_3$$ ### Step 3: Express the solution in parametric form Let $$x_2 = t$$ and $$x_3 = s$$, where $$t$$ and $$s$$ are free parameters. From the equation $$x_1 = 2x_2 - 2x_3$$, we have: $$x_1 = 2t - 2s$$ Thus, the solution vector $$\mathbf{x}$$ is: $$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 2t - 2s \\ t \\ s \end{bmatrix}$$ This can be rewritten as: $$\mathbf{x} = t \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$$ ### Step 4: Determine the basis The null space is spanned by the vectors: $$\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$$ Therefore, a basis for the null space of $$A$$ is: $$\left\{ \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \right\}$$ Would you like further details or clarification on this solution? Here are some related questions: 1. What is the rank of matrix $$A$$? 2. How do you find the nullity of a matrix? 3. What is the geometric interpretation of the null space of a matrix? 4. Can a matrix have a trivial null space? 5. How do you find the null space of larger matrices? **Tip**: The null space of a matrix represents all the solutions to the homogeneous system $$A\mathbf{x} = 0$$.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Null Space
Matrix Operations
Formulas
A * x = 0 (Homogeneous system of linear equations)
x_1 = 2x_2 - 2x_3
Theorems
Null Space Theorem: The null space of a matrix is the set of all solutions to A * x = 0.
Suitable Grade Level
Undergraduate Level (Linear Algebra)
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