JAMIA 2021 — Mathematics PYQ
JAMIA | Mathematics | 2021If A is a square matrix such that A2=I, then (A−I)3+(A−I)−7A is equal to
Choose the correct answer:
- A.
A
(Correct Answer) - B.
I−A
- C.
I+A
- D.
3A
A
Explanation
1. Expand the Cubes
Using the identity (a−b)3+(a+b)3=2a3+6ab2:
(A−I)3+(A+I)3=2A3+6AI2
(A−I)3+(A+I)3=2A3+6A
2. Apply the Condition A2=I
Since A2=I:
A3=A2⋅A=I⋅A=A
Substitute A3=A into the expanded expression:
2(A)+6A=8A
3. Final Simplification
Substitute this back into the full expression:
(A−I)3+(A+I)3−7A=8A−7A
(A−I)3+(A+I)3−7A=A
Final Result
A
Explanation
1. Expand the Cubes
Using the identity (a−b)3+(a+b)3=2a3+6ab2:
(A−I)3+(A+I)3=2A3+6AI2
(A−I)3+(A+I)3=2A3+6A
2. Apply the Condition A2=I
Since A2=I:
A3=A2⋅A=I⋅A=A
Substitute A3=A into the expanded expression:
2(A)+6A=8A
3. Final Simplification
Substitute this back into the full expression:
(A−I)3+(A+I)3−7A=8A−7A
(A−I)3+(A+I)3−7A=A
Final Result
A