Let \(J = \begin{bmatrix}1&1\\1&1\end{bmatrix}\), and let I and O be the identity and zero matrices of order 2.
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(J^3\) | (I) | \(I\) |
| (B) | \(J^5\) | (II) | \(16J\) |
| (C) | \(J^2 - 2J\) | (III) | \(O\) |
| (D) | \((J - I)^2\) | (IV) | \(4J\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
Multiplying row by column,
\( \displaystyle J^2 = \begin{bmatrix}1 + 1&1 + 1\\1 + 1&1 + 1\end{bmatrix} = \begin{bmatrix}2&2\\2&2\end{bmatrix} = 2J \)
Every higher power follows from this key relation \(J^2 = 2J\).
(A) \(J^3\)
This matches (IV) 4J.
\(J^3 = J^2 \cdot J = 2J^2 = 2(2J) = 4J\)
(B) \(J^5\)
This matches (II) 16J.
\(J^4 = J^3 \cdot J = 4J^2 = 8J\)
\(J^5 = J^4 \cdot J = 8J^2 = 16J\)
(C) \(J^2 - 2J\)
This matches (III) O.
\(J^2 - 2J = 2J - 2J = O\)
(D) \((J - I)^2\)
This matches (I) I. Since J and I commute,
\((J - I)^2 = J^2 - 2J + I\)
\(= 2J - 2J + I = I\)