To solve the problem provided, let’s analyze the given details and find a solution.

Given Information:

• A $3 \times 3$ matrix $A$ is given.
• The null space ($\text{Null}(A)$) of $A$ has a basis: $B = \left\{ \begin{bmatrix} 2 \\ -3 \\ 0 \end{bmatrix}, \begin{bmatrix} -4 \\ 2 \\ 0 \end{bmatrix} \right\}$
• We need to find a non-zero vector $\mathbf{v}$ in the row space of $A$ ($\text{Row}(A)$).

Key Observations:

1. Since the null space of $A$ is spanned by the given vectors, the dimension of $\text{Null}(A)$ is 2.

2. By the Rank-Nullity Theorem: $\text{dim}(\text{Row}(A)) + \text{dim}(\text{Null}(A)) = \text{number of columns of } A.$ For $A$, the number of columns is 3. Therefore: $\text{dim}(\text{Row}(A)) = 3 - 2 = 1.$ Thus, the row space is one-dimensional, and any vector in the row space will be a scalar multiple of a single row vector.

3. Any vector $\mathbf{v}$ in $\text{Row}(A)$ will satisfy: $\mathbf{v} \cdot \mathbf{n} = 0,$ where $\mathbf{n}$ is any vector in $\text{Null}(A)$. This is because $\text{Null}(A)$ is orthogonal to $\text{Row}(A)$.

Approach:

To find a non-zero vector $\mathbf{v} \in \text{Row}(A)$:

1. Let $\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}$.
2. Enforce the orthogonality condition for the null space basis vectors: $\mathbf{v} \cdot \begin{bmatrix} 2 \\ -3 \\ 0 \end{bmatrix} = 0 \quad \text{and} \quad \mathbf{v} \cdot \begin{bmatrix} -4 \\ 2 \\ 0 \end{bmatrix} = 0.$
3. These conditions give two equations: $2v_1 - 3v_2 + 0v_3 = 0 \quad \text{(1)}$ $-4v_1 + 2v_2 + 0v_3 = 0 \quad \text{(2)}.$

Solve the System:

From equation (1): $2v_1 - 3v_2 = 0 \implies v_1 = \frac{3}{2}v_2.$

From equation (2): $-4v_1 + 2v_2 = 0 \implies -4\left(\frac{3}{2}v_2\right) + 2v_2 = 0.$ Simplify: $-6v_2 + 2v_2 = 0 \implies -4v_2 = 0 \implies v_2 = 0.$

Substitute $v_2 = 0$ into $v_1 = \frac{3}{2}v_2$: $v_1 = 0.$

Thus, $v_3$ can be any non-zero value. Choose $v_3 = 1$, and the vector becomes: $\mathbf{v} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}.$

Final Answer:

A non-zero vector $\mathbf{v} \in \text{Row}(A)$ is: $\mathbf{v} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}.$

Would you like detailed verification or have further questions? Here are 5 related concepts to explore:

1. What is the Rank-Nullity Theorem and how is it applied?
2. How do we interpret the null space geometrically?
3. How do orthogonality properties relate to the row and null spaces?
4. What is the basis of the column space of a matrix?
5. Can you find the rank of $A$ based on its null space?

Tip: Always verify the orthogonality of your solution when working with row and null spaces!