A machine costing Rs.2,00,000 has a useful life of 5 years.The estimated scrap value is Rs.20,000. By using straight line method, the annual…

Given: $AB = A$AB=A and $BA = B$BA=B, where $A$A and $B$B are square matrices of the same order.

---

Multiply $AB = A$AB=A by $A$A on the right:

$$ABA = A^2$$ABA=A2

Rewrite the left side:

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

$$ABA = A \cdot B$$ABA=A⋅B

$$ABA = AB$$ABA=AB

$$ABA = A$$ABA=A

Therefore:

$$A^2 = A$$A2=A

---

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

$$A^3 = A^2 \cdot A$$A3=A2⋅A

$$A^3 = A \cdot A$$A3=A⋅A

$$A^3 = A$$A3=A

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

For any positive integer $n$n:

$$A^n = A$$An=A

Therefore:

$$A^{2024} = A$$A2024=A

---

Multiply $BA = B$BA=B by $B$B on the right:

$$BAB = B^2$$BAB=B2

Rewrite the left side:

$$BAB = (BA) \cdot B$$BAB=(BA)⋅B

$$BAB = B \cdot B$$BAB=B⋅B

$$BAB = B^2$$BAB=B2

Also:

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

$$BAB = B \cdot A$$BAB=B⋅A

$$BAB = BA$$BAB=BA

$$BAB = B$$BAB=B

Therefore:

$$B^2 = B$$B2=B

---

By the same reasoning as for $A$A:

$$B^{2024} = B$$B2024=B

---

The value of $A^{2024} + B^{2024}$A2024+B2024:

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

Therefore, the answer is $A + B$A+B.