Question

Match the LIST-I with LIST-II

Choose the correct answer from the options given below:

• A. A-II, B-IV, C-III, D-I
• B. A-IV, B-II, C-I, D-III
• C. A-II, B-IV, C-I, D-III
• D. A-IV, B-II, C-III, D-I

Hint:

To quickly find the standard matrix of $T(x_1, x_2)$, simply read the coefficients of $x_1$ and $x_2$ along each row!

Answer 1

The Correct Option is C

Step 1 : Concept:
This question involves constructing matrix representations $[T]_B$ of linear operators $T: \mathbb{R}^2 \to \mathbb{R}^2$ relative to the standard basis $B = \{(1,0), (0,1)\}$.

Step 2 : Key Formulas and Approach:
For standard basis $e_1 = (1,0)$ and $e_2 = (0,1)$, the matrix representation is: \[ [T]_B = \begin{bmatrix} | & | T(e_1) & T(e_2) | & | \end{bmatrix} \]

Step 3 : Step-by-step Explanation:

• Item A: $T(x_1, x_2) = (x_1, 0)$ $T(e_1) = T(1,0) = (1,0) = 1e_1 + 0e_2$ $T(e_2) = T(0,1) = (0,0) = 0e_1 + 0e_2$ $[T]_B = \begin{bmatrix} 1 & 0
0 & 0 \end{bmatrix}$. Matches with II.

• Item B: $T(x_1, x_2) = (x_1 - x_2, x_1 + x_2)$ $T(e_1) = T(1,0) = (1,1) = 1e_1 + 1e_2$ $T(e_2) = T(0,1) = (-1,1) = -1e_1 + 1e_2$ $[T]_B = \begin{bmatrix} 1 & -1
1 & 1 \end{bmatrix}$. Matches with IV.

• Item C: $T(x_1, x_2) = (0, x_2)$ $T(e_1) = T(1,0) = (0,0) = 0e_1 + 0e_2$ $T(e_2) = T(0,1) = (0,1) = 0e_1 + 1e_2$ $[T]_B = \begin{bmatrix} 0 & 0
0 & 1 \end{bmatrix}$. Matches with I.

• Item D: $T(x_1, x_2) = (x_1, x_2)$ (Identity operator) $T(e_1) = (1,0)$ $T(e_2) = (0,1)$ $[T]_B = \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix}$. Matches with III.

Step 4 : Final Answer:
The correct matching is A-II, B-IV, C-I, D-III, which corresponds to option (C).