Let P and Q be `3xx3` matrices such that `P!=Q`. If `P^(3)=Q^(3)` and `P^(2)Q=Q^(2)P` then determinant of `(P^(2)+Q^(2))` is equal to

Options

A. `-2`
B. 0.01
C. 0
D. `-1`

Answer

To solve the problem, we need to analyze the given conditions involving the matrices \( P \) and \( Q \).

### Step-by-Step Solution:

1. **Given Conditions**: We have two matrices \( P \) and \( Q \) such that:
- \( P^3 = Q^3 \) (Equation 1)
- \( P^2 Q = Q^2 P \) (Equation 2)
- \( P \neq Q \)

2. **Rearranging Equation 2**: From Equation 2, we can rearrange it as follows:
\[ P^2 Q - Q^2 P = 0 \]
This can be factored as:
\[ P^2 Q - Q P^2 = 0 \implies (P^2 - Q^2)Q = 0 \]

3. **Factoring**: We can factor \( P^2 - Q^2 \) using the difference of squares:
\[ P^2 - Q^2 = (P - Q)(P + Q) \]
Therefore, we can rewrite our equation:
\[ (P - Q)(P + Q)Q = 0 \]

4. **Analyzing the Factors**: Since \( P \neq Q \), we know \( P - Q \neq 0 \). Thus, for the product to equal zero, we must have:
\[ (P + Q)Q = 0 \]
This implies that either \( P + Q = 0 \) or \( Q = 0 \). However, since \( P \) and \( Q \) are matrices and we are not given that they are zero matrices, we focus on \( P + Q = 0 \).

5. **Using Equation 1**: From \( P^3 = Q^3 \), we can also express this in terms of \( P \) and \( Q \):
\[ P^3 - Q^3 = 0 \implies (P - Q)(P^2 + PQ + Q^2) = 0 \]
Since \( P \neq Q \), it follows that:
\[ P^2 + PQ + Q^2 = 0 \]

6. **Finding \( P^2 + Q^2 \)**: We know that:
\[ P^2 + Q^2 = -(PQ) \]

7. **Determinant of \( P^2 + Q^2 \)**: We need to find the determinant of \( P^2 + Q^2 \):
\[ \det(P^2 + Q^2) = \det(-PQ) = (-1)^3 \det(PQ) = -\det(PQ) \]
However, since \( P^2 + Q^2 \) is derived from the conditions, we need to analyze further.

8. **Conclusion**: If \( P^2 + Q^2 \) is invertible, then its determinant cannot be zero. However, since we derived that \( P^2 + Q^2 = -(PQ) \) and we have \( P \neq Q \), we conclude:
\[ \det(P^2 + Q^2) = 0 \]

### Final Answer:
The determinant of \( P^2 + Q^2 \) is equal to \( 0 \).