Let \(A = \begin{bmatrix}2&4\\-2&6\end{bmatrix}\) and let I be the identity matrix of order 2.
Match each condition on the matrix X in List-I with the matrix X in List-II
| List-I (Condition) | List-II (X) | ||
|---|---|---|---|
| (A) | \(X + A = 2I\) | (I) | \(\begin{bmatrix}0&3\\-3&0\end{bmatrix}\) |
| (B) | \(2X = A\) | (II) | \(\begin{bmatrix}2&-2\\4&6\end{bmatrix}\) |
| (C) | \(X = A'\) | (III) | \(\begin{bmatrix}0&-4\\2&-4\end{bmatrix}\) |
| (D) | \(2X = A - A'\) | (IV) | \(\begin{bmatrix}1&2\\-1&3\end{bmatrix}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Each X is found with the rules of addition, scalar multiplication and transpose, element by element.
(A) \(X + A = 2I\)
This matches (III) \(\begin{bmatrix}0&-4\\2&-4\end{bmatrix}\).
\( \displaystyle X = 2I - A = \begin{bmatrix}2 - 2&0 - 4\\0 + 2&2 - 6\end{bmatrix} \)
(B) \(2X = A\)
This matches (IV) \(\begin{bmatrix}1&2\\-1&3\end{bmatrix}\).
\( \displaystyle X = \frac{1}{2}A = \begin{bmatrix}1&2\\-1&3\end{bmatrix} \)
(C) \(X = A'\)
This matches (II) \(\begin{bmatrix}2&-2\\4&6\end{bmatrix}\). The rows of A become the columns.
(D) \(2X = A - A'\)
This matches (I) \(\begin{bmatrix}0&3\\-3&0\end{bmatrix}\).
\( \displaystyle A - A' = \begin{bmatrix}0&6\\-6&0\end{bmatrix} \)
\( \displaystyle X = \frac{1}{2}(A - A') = \begin{bmatrix}0&3\\-3&0\end{bmatrix} \)