Question

Define \( T : \mathbb{R}^3 \to \mathbb{R}^3 \) by \[ T(x, y, z) = (x + z, 2x + 3y + 5z, 2y + 2z), \quad \text{for all } (x, y, z) \in \mathbb{R}^3 \] Then, which one of the following is TRUE?

• A. \( T \) is one-one and \( T \) is NOT onto
• B. \( T \) is NOT one-one and \( T \) is onto
• C. \( T \) is one-one and \( T \) is onto
• D. \( T \) is NOT one-one and \( T \) is NOT onto

Hint:

To check if a linear transformation is one-to-one, solve the system \( T(x_1, y_1, z_1) = T(x_2, y_2, z_2) \) and show that it implies \( (x_1, y_1, z_1) = (x_2, y_2, z_2) \). For onto, check if the transformation maps to the entire codomain.

Answer 1

The Correct Option is D

Step 1: Check if \( T \) is one-to-one.
To check if \( T \) is one-to-one, we solve \( T(x_1, y_1, z_1) = T(x_2, y_2, z_2) \). This implies: \[ x_1 + z_1 = x_2 + z_2, \quad 2x_1 + 3y_1 + 5z_1 = 2x_2 + 3y_2 + 5z_2, \quad 2y_1 + 2z_1 = 2y_2 + 2z_2 \] By solving this system, we find that \( (x_1, y_1, z_1) = (x_2, y_2, z_2) \), so \( T \) is one-to-one. Step 2: Check if \( T \) is onto.
Since \( T \) is a linear transformation and the coefficient matrix is invertible, \( T \) is onto. Final Answer: \[ \boxed{T \text{ is one-one and onto}} \]