Discussion
Let \(T : \mathbf{R}^{3} \to \mathbf{R}^{2}\) be a linear map such that :
\[ T\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad T\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]
and
\[ T\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}. \]
Then \(T\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} =\)
Answer: A
\(T(1, 1, 1)\) is \(\begin{bmatrix} 2 \\ 1 \end{bmatrix}\).
Write the vector as a combination of the vectors whose images are given:
\[ (1, 1, 1) = (1, 0, 0) + (0, 1, 1) \]
By linearity,
\[ T(1, 1, 1) = (1, 0) + (1, 1) = (2, 1) \]
The information \(T(0, 1, 0) = (0, 1)\) is not needed. Adding all three given images gives (2, 2), which is the trap in option D; (1, 1) is the image of (0, 1, 1) alone.
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