If A and B are square matrices of same order such that, \( \mathrm{AB=BA,\ A^{-1}=A^2} \) and \( \mathrm{B^{-1}=B^3} \), then the value of \( \mathrm{(AB)}^{2040} \), is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Identity matrix
Note: The given answer is as per the official NTA Answer Key. However, more than one answer is possible: Option (1), Option (2) and Option (4) are all correct, as explained below.
Multiplying \( A^{-1} = A^2 \) by A, and \( B^{-1} = B^3 \) by B:
\( A^3 = I,\qquad B^4 = I \)
Since AB = BA, \( (AB)^{2040} = A^{2040}B^{2040} \), and 2040 = 3 × 680 = 4 × 510:
\( (AB)^{2040} = (A^3)^{680}(B^4)^{510} = I \cdot I = I \)
Option (1) Diagonal matrix — Should also be correct, though not the answer given by NTA. Every entry of I off the main diagonal is 0, so I is a diagonal matrix.
Option (2) Identity matrix — Correct. As shown above, \( (AB)^{2040} = I \).
Option (3) Null matrix — Incorrect. The diagonal entries of I are 1, not 0.
Option (4) Scalar matrix — Should also be correct, though not the answer given by NTA. \( I = 1\cdot I \), a diagonal matrix with equal diagonal entries, so it is a scalar matrix.
Advice for the exam: When more than one option is correct, mark the most specific one, here “Identity matrix”, which already implies the other two. If NTA’s provisional answer key marks a different option, challenge it during the answer-key challenge window, giving proper reasoning. NTA does revise its key when a challenge is well justified, so a sound challenge can raise your marks.