Discussion
\(T_{1}\), \(T_{2}\) and \(T_{3}\) are three maps defined on \(\mathbf{R}^{3}\) as :
\[ T_{1}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x + 1 \\ y \\ z \end{bmatrix}, \]
\[ T_{2}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} xy \\ z \end{bmatrix} \]
\[ T_{3}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x + y \\ y + z \\ z + x \end{bmatrix}. \]
Which of these maps are linear ?
Answer: C
Only \(T_{3}\) is linear.
\(T_{3}\) sends \((x, y, z)\) to \((x + y, y + z, z + x)\); each component is a linear combination of \(x, y, z\), so it is the linear map given by a 3 by 3 matrix.
\(T_{1}\) is a translation: \(T_{1}(0) = (1, 0, 0) \ne 0\), and a linear map must send 0 to 0. \(T_{2}\) contains the product \(xy\): doubling the input multiplies the first component by 4, not 2, so it is not homogeneous. Any option listing \(T_{1}\) or \(T_{2}\) is wrong.
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