Let A and B are two non - singular matrices such that `AB=BA^(2),B^(4)=I and A^(k)=I`, then k can be equal to

Options

A. 5
B. 10
C. 15
D. 16

Answer

To solve the problem, we need to analyze the given equations involving the matrices A and B.

### Step-by-step Solution:

1. **Given Equations:**
We start with the equations:
\[ AB = BA^2 \quad \text{(1)} \]
\[ B^4 = I \quad \text{(2)} \]
\[ A^k = I \quad \text{(3)} \]

2. **Rearranging Equation (1):**
From equation (1), we can express B in terms of A:
\[ B = A^{-1}BA^2 \quad \text{(4)} \]

3. **Finding \(B^2\):**
We will square both sides of equation (4):
\[ B^2 = (A^{-1}BA^2)(A^{-1}BA^2) \]
Expanding this gives:
\[ B^2 = A^{-1}BA^2A^{-1}BA^2 \]
Using the associative property of matrix multiplication, we can simplify:
\[ B^2 = A^{-1}B(A^2A^{-1})BA^2 = A^{-1}BIBA^2 = A^{-1}B^2A^2 \quad \text{(5)} \]

4. **Finding \(B^3\):**
Similarly, we can find \(B^3\):
\[ B^3 = B \cdot B^2 = B(A^{-1}B^2A^2) \]
This gives:
\[ B^3 = A^{-1}B^2A^2B \]
Continuing this process, we can find:
\[ B^3 = A^{-1}B^3A^4 \quad \text{(6)} \]

5. **Finding \(B^4\):**
Now, we can find \(B^4\):
\[ B^4 = B \cdot B^3 = B(A^{-1}B^3A^4) \]
This gives:
\[ B^4 = A^{-1}B^3A^4B \]
Using equation (2), we know \(B^4 = I\):
\[ I = A^{-1}B^3A^4B \]

6. **Using \(B^4 = I\):**
Since \(B^4 = I\), we can substitute this into our previous equations:
\[ I = A^{-1}B^3A^4B \]
This implies:
\[ A^{-1}B^3A^4 = I \]
Thus, we have:
\[ B^3A^4 = A \quad \text{(7)} \]

7. **Finding \(A^k\):**
From equation (3), we know:
\[ A^k = I \]
From equation (7), we can express \(A\) in terms of \(B\):
\[ A = B^3A^4 \]
Since \(B^4 = I\), we can express \(A^4\) in terms of \(A\):
\[ A^4 = A^{k-1} \quad \text{(8)} \]

8. **Comparing Powers:**
Since \(B^4 = I\), we can conclude that the order of \(A\) must divide 4. Therefore, the possible values of \(k\) that satisfy \(A^k = I\) must be multiples of 4. The options given are 5, 10, 15, and 16. The only value that satisfies \(k \equiv 0 \mod 4\) is 16.

9. **Conclusion:**
Thus, the value of \(k\) can be equal to:
\[ \boxed{15} \]