If A and B are two square matrices of order `3xx3` which satify `AB=A` and `BA=B`, then
`(A+I)^(5)` is equal to (where I is idensity matric)

Options

A. `I+60I`
B. `I+16A`
C. `I+31A`
D. none of these

Explanation

To solve the problem, we need to find the value of \((A + I)^5\) given the conditions \(AB = A\) and \(BA = B\).

### Step-by-Step Solution:

1. **Understanding the Given Conditions**:
We have two square matrices \(A\) and \(B\) of order \(3 \times 3\) such that:
\[ AB = A \quad \text{and} \quad BA = B \]
From \(AB = A\), we can rearrange it to get:
\[ A(B - I) = 0 \]
This implies that either \(A = 0\) or \(B = I\) (the identity matrix).

2. **Using the Second Condition**:
From \(BA = B\), we can rearrange it to get:
\[ (B - I)A = 0 \]
This implies that either \(B = 0\) or \(A = I\).

3. **Conclusion from Conditions**:
The conditions \(AB = A\) and \(BA = B\) suggest that both \(A\) and \(B\) are idempotent matrices. Specifically, we can conclude that:
\[ A^2 = A \quad \text{and} \quad B^2 = B \]

4. **Expanding \((A + I)^5\)**:
We can use the binomial theorem to expand \((A + I)^5\):
\[ (A + I)^5 = \sum_{k=0}^{5} \binom{5}{k} A^k I^{5-k} \]
Since \(I^n = I\) for any positive integer \(n\), we have:
\[ (A + I)^5 = \binom{5}{0} I^5 + \binom{5}{1} A I^4 + \binom{5}{2} A^2 I^3 + \binom{5}{3} A^3 I^2 + \binom{5}{4} A^4 I + \binom{5}{5} A^5 \]
Simplifying this, we get:
\[ (A + I)^5 = I + 5A + 10A^2 + 10A^3 + 5A^4 + A^5 \]

5. **Using Idempotent Property**:
Since \(A^2 = A\), we can replace higher powers of \(A\):
\[ A^3 = A, \quad A^4 = A, \quad A^5 = A \]
Thus, substituting these into our expansion:
\[ (A + I)^5 = I + 5A + 10A + 10A + 5A + A \]
This simplifies to:
\[ (A + I)^5 = I + (5 + 10 + 10 + 5 + 1)A = I + 31A \]

6. **Final Result**:
Therefore, the expression \((A + I)^5\) simplifies to:
\[ (A + I)^5 = 31A + I \]

### Final Answer:
\[ (A + I)^5 = 31A + I \]