This question involves finding the null space of a matrix $A$ and determining the dimension of the null space, $\dim(\text{Null}(A))$.

Given Matrix $A$:

$A = \begin{pmatrix} 1 & 5 & 3 \\ 2 & 5 & 1 \end{pmatrix}$

The null space of a matrix $A$, denoted as $\text{Null}(A)$, consists of all vectors $\mathbf{x}$ such that: $A\mathbf{x} = \mathbf{0}$ This means we need to find the solutions to the system of linear equations represented by $A\mathbf{x} = 0$.

Step-by-step explanation:

1. System of Equations:

Multiplying the matrix $A$ by the vector $\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$, we get the system of equations: $\begin{pmatrix} 1 & 5 & 3 \\ 2 & 5 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ This yields the following two equations:

1. $x_1 + 5x_2 + 3x_3 = 0$
2. $2x_1 + 5x_2 + x_3 = 0$

2. Solving the System:

We can solve this system using Gaussian elimination.

First, subtract 2 times the first equation from the second equation to eliminate $x_1$: $(2x_1 + 5x_2 + x_3) - 2(x_1 + 5x_2 + 3x_3) = 0$ This simplifies to: $-9x_3 = 0 \quad \Rightarrow \quad x_3 = 0$

Now substitute $x_3 = 0$ into the first equation: $x_1 + 5x_2 = 0 \quad \Rightarrow \quad x_1 = -5x_2$

3. General Solution:

So, the solution to the system can be written as: $x_1 = -5x_2, \quad x_3 = 0$ Thus, the general solution for the null space is: $\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} -5x_2 \\ x_2 \\ 0 \end{pmatrix} = x_2 \begin{pmatrix} -5 \\ 1 \\ 0 \end{pmatrix}$ This means the null space is spanned by the vector $\begin{pmatrix} -5 \\ 1 \\ 0 \end{pmatrix}$.

4. Dimension of the Null Space:

Since the null space is spanned by a single vector, the dimension of the null space is 1. This is written as: $\dim(\text{Null}(A)) = 1$

Answer Choice:

The correct answer is: $\text{Null}(A) = \text{span} \left\{ \begin{pmatrix} -5 \\ 1 \\ 0 \end{pmatrix} \right\}, \quad \dim(\text{Null}(A)) = 1$ Looking at the options, this matches the third option: $\text{Null}(A) = \text{span} \left\{ \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \right\}, \dim(\text{Null}(A)) = 1$ (Up to scalar multiples, $\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$ is a scalar multiple of $\begin{pmatrix} -5 \\ 1 \\ 0 \end{pmatrix}$, so these are equivalent.)

Would you like more details, or do you have any further questions?

Here are five related questions to expand your understanding:

1. What is the geometric interpretation of the null space of a matrix?
2. How is the rank of a matrix related to its null space?
3. What does it mean when the dimension of the null space is zero?
4. How do you find the null space of a larger matrix (e.g., a 3x4 matrix)?
5. Can the null space of a matrix ever be larger than the matrix's row space?

Tip: Always check the rank of a matrix to help determine the dimension of its null space. This is known as the Rank-Nullity Theorem.