Let P and Q be `3xx3` matrices with `P!=Q` . If `P^3=""Q^3a n d""P^2Q""=""Q^2P` , then determinant of `(P^2+""Q^2)` is equal to (1) `2` (2) 1 (3) 0 (4) `1`

Options

A. `-2`
B. 1
C. 0
D. `-1`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write down the given equations
We are given two equations:
1. \( P^3 = Q^3 \)
2. \( P^2 Q = Q^2 P \)

### Step 2: Rearrange the first equation
From the first equation, we can rearrange it as:
\[ P^3 - Q^3 = 0 \]
This can be factored using the difference of cubes:
\[ (P - Q)(P^2 + PQ + Q^2) = 0 \]

### Step 3: Analyze the factored equation
The equation \( (P - Q)(P^2 + PQ + Q^2) = 0 \) implies that either:
1. \( P - Q = 0 \) (which means \( P = Q \)), or
2. \( P^2 + PQ + Q^2 = 0 \)

Since we are given that \( P \neq Q \), we discard the first possibility. Therefore, we have:
\[ P^2 + PQ + Q^2 = 0 \]

### Step 4: Rearrange the second equation
From the second equation \( P^2 Q = Q^2 P \), we can rearrange it as:
\[ P^2 Q - Q^2 P = 0 \]
Factoring gives:
\[ P^2 Q - Q P Q = 0 \implies (P^2 - Q^2)Q = 0 \]

### Step 5: Analyze the second factored equation
This implies that either:
1. \( P^2 - Q^2 = 0 \) (which means \( P^2 = Q^2 \)), or
2. \( Q = 0 \)

Since \( P \) and \( Q \) are not zero matrices, we consider \( P^2 = Q^2 \).

### Step 6: Combine the results
From \( P^2 + PQ + Q^2 = 0 \) and \( P^2 = Q^2 \), we substitute \( Q^2 \) into the first equation:
\[ P^2 + PQ + P^2 = 0 \implies 2P^2 + PQ = 0 \]

### Step 7: Solve for \( P^2 + Q^2 \)
Since \( P^2 = Q^2 \), we can write:
\[ P^2 + Q^2 = 2P^2 \]

### Step 8: Find the determinant
Now, we need to find the determinant:
\[ \text{det}(P^2 + Q^2) = \text{det}(2P^2) = 2^3 \text{det}(P^2) = 8 \text{det}(P^2) \]

### Step 9: Determine the value of \( \text{det}(P^2) \)
Since \( P^2 + Q^2 = 0 \) implies that \( P^2 \) is a null matrix, we have:
\[ \text{det}(P^2) = 0 \]

### Final Result
Thus, we find:
\[ \text{det}(P^2 + Q^2) = 0 \]

### Conclusion
The determinant of \( P^2 + Q^2 \) is equal to **0**.