Let A be a `3xx3` matrix and `S={((x),(y),(z))x,y,z, in R}`
Define `f: S rarr S ` by
`f((x),(y),(z))=A((x),(y),(z))`
Suppose `f((x),(y),(z))=((0),(0),(0)) implies x=y=z=0` Then

Options

A. f is one - to - one
B. f cannot be onto
C. A is not invertible
D. A=O

Answer

To solve the problem, we need to analyze the function \( f: S \to S \) defined by \( f((x), (y), (z)) = A((x), (y), (z)) \), where \( A \) is a \( 3 \times 3 \) matrix and \( S = \{(x), (y), (z) \mid x, y, z \in \mathbb{R}\} \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f \) takes a vector \( (x, y, z) \) from set \( S \) and maps it to another vector in \( S \) by multiplying it with the matrix \( A \).

2. **Condition Given**:
We are given that \( f((x), (y), (z)) = ((0), (0), (0)) \) implies \( x = y = z = 0 \). This means that the only solution to the equation \( A((x), (y), (z)) = ((0), (0), (0)) \) is the zero vector.

3. **Implication of the Condition**:
The condition implies that the kernel (null space) of the matrix \( A \) contains only the zero vector. This means that \( A \) is injective (one-to-one).

4. **Injectivity**:
To show that \( f \) is injective, assume \( f((x_1), (y_1), (z_1)) = f((x_2), (y_2), (z_2)) \). This leads to:
\[ A((x_1), (y_1), (z_1)) = A((x_2), (y_2), (z_2)) \]
By the property of linear transformations, this implies:
\[ A((x_1 - x_2), (y_1 - y_2), (z_1 - z_2)) = ((0), (0), (0)) \]
Given our condition, this means:
\[ (x_1 - x_2) = 0, \quad (y_1 - y_2) = 0, \quad (z_1 - z_2) = 0 \]
Thus, \( x_1 = x_2 \), \( y_1 = y_2 \), \( z_1 = z_2 \). Therefore, \( f \) is injective.

5. **Surjectivity**:
To check if \( f \) is surjective (onto), we need to determine if for every \( (a, b, c) \in S \), there exists \( (x, y, z) \in S \) such that:
\[ A((x), (y), (z)) = ((a), (b), (c)) \]
This requires \( A \) to be invertible. If \( A \) is not invertible, then there may exist vectors in \( S \) that cannot be expressed as \( A((x), (y), (z)) \). Since we are not given that \( A \) is invertible, we cannot conclude that \( f \) is surjective.

6. **Conclusion**:
Since \( f \) is injective but not necessarily surjective, we conclude that \( f \) is a one-to-one function but not onto.

### Final Answer:
The function \( f \) is one-to-one (injective) but not onto (surjective).