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The question states that matrix \( A \) is a \( 2 \times 3 \) matrix in row-reduced echelon form (RREF), and vector \( v = \begin{pmatrix} 1 \\ -2 \\ -1 \end{pmatrix} \) is a solution to the homogeneous equation \( Av = 0 \).

We are tasked with identifying which of the three matrices given could be equal to \( A \).

### Step-by-step approach:

1. Since \( A \) is in RREF and \( v \) is a solution to \( Av = 0 \), when we multiply the matrix \( A \) by the vector \( v \), we must get the zero vector.
   
   The matrix \( A \) is \( 2 \times 3 \), and the vector \( v \) has 3 components. This confirms the multiplication \( Av = 0 \) is valid.

2. **Matrix A options**:
   
   (i) \( A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -2 \end{pmatrix} \)
   
   (ii) \( A = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \end{pmatrix} \)
   
   (iii) \( A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \end{pmatrix} \)

3. **Multiplying each matrix by \( v = \begin{pmatrix} 1 \\ -2 \\ -1 \end{pmatrix} \)** to check if \( Av = 0 \) holds.

### Testing Option (i):
\[
A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -2 \end{pmatrix}, \quad v = \begin{pmatrix} 1 \\ -2 \\ -1 \end{pmatrix}
\]
\[
Av = \begin{pmatrix} 1 \cdot 1 + 0 \cdot (-2) + 1 \cdot (-1) \\ 0 \cdot 1 + 1 \cdot (-2) + (-2) \cdot (-1) \end{pmatrix} = \begin{pmatrix} 1 - 1 \\ -2 + 2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]
This is correct.

### Testing Option (ii):
\[
A = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \end{pmatrix}, \quad v = \begin{pmatrix} 1 \\ -2 \\ -1 \end{pmatrix}
\]
\[
Av = \begin{pmatrix} 1 \cdot 1 + 0 \cdot (-2) + (-1) \cdot (-1) \\ 0 \cdot 1 + 1 \cdot (-2) + 2 \cdot (-1) \end{pmatrix} = \begin{pmatrix} 1 + 1 \\ -2 - 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \end{pmatrix}
\]
This is incorrect.

### Testing Option (iii):
\[
A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \end{pmatrix}, \quad v = \begin{pmatrix} 1 \\ -2 \\ -1 \end{pmatrix}
\]
\[
Av = \begin{pmatrix} 1 \cdot 1 + 0 \cdot (-2) + 1 \cdot (-1) \\ 0 \cdot 1 + 1 \cdot (-2) + 2 \cdot (-1) \end{pmatrix} = \begin{pmatrix} 1 - 1 \\ -2 - 2 \end{pmatrix} = \begin{pmatrix} 0 \\ -4 \end{pmatrix}
\]
This is incorrect.

### Conclusion:
The correct matrix for \( A \) is **Option (i)**: \( \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -2 \end{pmatrix} \).

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Would you like further details on any of the steps? Here are some related questions:

1. How is the row-reduced echelon form (RREF) of a matrix determined?
2. What is the geometric interpretation of the solution to a homogeneous system \( Av = 0 \)?
3. How do we compute matrix-vector multiplication?
4. Why is the vector \( v \) called a solution to \( Av = 0 \)?
5. What are the key properties of matrices in RREF?

**Tip:** Always check if a matrix is in row-reduced echelon form by verifying if each leading entry is 1, and all entries above and below the leading entry in the column are zeros.

A is a 2x3 matrix in RREF, and vector v is a solution to the homogeneous equation Av = 0, where v = (1, -2, -1). Which of the following matrices could A be equal to?

Explore how to determine which matrix in RREF form satisfies the equation Av = 0, where v = (1, -2, -1). This problem involves matrix-vector multiplication and understanding the properties of row-reduced echelon form (RREF) for college-level linear algebra.

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