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If A and B are two square matrices of same order such that AB=AAB = A and BA=BBA = B, then the value of A2024+B2024A^{2024} + B^{2024} is equal to

Solution

✅ Correct Option: 3

Given: AB=AAB = A and BA=BBA = B, where AA and BB are square matrices of the same order.


Multiply AB=AAB = A by AA on the right:

ABA=A2ABA = A^2

Rewrite the left side:

ABA=A(BA)ABA = A(BA)

ABA=A⋅BABA = A \cdot B

ABA=ABABA = AB

ABA=AABA = A

Therefore:

A2=AA^2 = A


Since A2=AA^2 = A, all higher powers of AA equal AA:

A3=A2⋅AA^3 = A^2 \cdot A

A3=A⋅AA^3 = A \cdot A

A3=AA^3 = A

Similarly, A4=AA^4 = A, A5=AA^5 = A, and so on.

For any positive integer nn:

An=AA^n = A

Therefore:

A2024=AA^{2024} = A


Multiply BA=BBA = B by BB on the right:

BAB=B2BAB = B^2

Rewrite the left side:

BAB=(BA)⋅BBAB = (BA) \cdot B

BAB=B⋅BBAB = B \cdot B

BAB=B2BAB = B^2

Also:

BAB=B(AB)BAB = B(AB)

BAB=B⋅ABAB = B \cdot A

BAB=BABAB = BA

BAB=BBAB = B

Therefore:

B2=BB^2 = B


By the same reasoning as for AA:

B2024=BB^{2024} = B


The value of A2024+B2024A^{2024} + B^{2024}:

A2024+B2024=A+BA^{2024} + B^{2024} = A + B

Therefore, the answer is A+BA + B.

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