Discussion
Let T be a linear operator on \(\mathbf{R}^{3}\), given by :
\[ T\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2x + 2y + z \\ 2y + z \\ 2x + 3y + z \end{bmatrix} \]
Then the matrix representation of T with respect to standard basis of \(\mathbf{R}^{3}\) is :
Answer: A
The matrix is \(\begin{bmatrix} 2 & 2 & 1 \\ 0 & 2 & 1 \\ 2 & 3 & 1 \end{bmatrix}\).
The columns are the images of the standard basis vectors:
\[ \begin{aligned} T(e_{1}) &= (2, 0, 2) \\ T(e_{2}) &= (2, 2, 3) \\ T(e_{3}) &= (1, 1, 1) \end{aligned} \]
Placed as columns, they give rows (2, 2, 1), (0, 2, 1), (2, 3, 1), which are just the coefficients of each output component. Option C is the transpose, with the images written as rows, and the other two rearrange the entries.
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