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The problem is asking to prove that \(\det(P) = \det(A)\det(D)\) for the given block matrix \(P\), where \(P\) is defined as:

\[
P = \begin{pmatrix}
a_{11} & a_{12} & 0 & 0 \\
a_{21} & a_{22} & 0 & 0 \\
b_{11} & b_{12} & d_{11} & d_{12} \\
b_{21} & b_{22} & d_{21} & d_{22}
\end{pmatrix}
\]

The submatrices are:

\[
A = \begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{pmatrix}, \quad B = \begin{pmatrix}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{pmatrix}, \quad D = \begin{pmatrix}
d_{11} & d_{12} \\
d_{21} & d_{22}
\end{pmatrix}
\]

The matrix \(P\) is a block matrix, and it is structured in such a way that you can apply the determinant formula for block matrices.

### Determinant of a Block Matrix

For a block matrix of the form:

\[
\begin{pmatrix}
A & 0 \\
B & D
\end{pmatrix}
\]

where \(A\) and \(D\) are square matrices and \(0\) is a zero matrix, the determinant is given by:

\[
\det\left(\begin{pmatrix}
A & 0 \\
B & D
\end{pmatrix}\right) = \det(A) \cdot \det(D)
\]

This result holds because the determinant of a block upper (or lower) triangular matrix is the product of the determinants of the diagonal blocks.

### Application to the Problem

Given the matrix \(P\), we see that it has the form:

\[
P = \begin{pmatrix}
A & 0 \\
B & D
\end{pmatrix}
\]

So, by the determinant formula for block matrices, we have:

\[
\det(P) = \det(A) \cdot \det(D)
\]

### Conclusion

This directly shows that:

\[
\boxed{\det(P) = \det(A)\det(D)}
\]

This is the required proof.

---

Would you like further details or explanations on any part of this proof? 

Here are five follow-up questions you might consider:
1. What are some other types of block matrices where the determinant can be simplified?
2. How would the proof change if \(P\) were not a block upper triangular matrix?
3. Can this determinant property be extended to larger block matrices?
4. What role do the submatrices \(B\) and \(0\) play in the determinant of \(P\)?
5. How does this property of determinants simplify calculations in linear algebra?

**Tip:** When dealing with block matrices, identifying the structure (upper triangular, lower triangular, etc.) is crucial as it often simplifies determinant calculations.

Explore the proof that shows \(\det(P) = \det(A)\det(D)\) for a block matrix \(P\), where \(P\) is structured as defined. This detailed explanation covers the determinant formula for block matrices, providing insights into linear algebra concepts and determinant properties.

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