The product of a `9xx4` matrix and a `4xx9` matrix contains a variable x in exactly two places. If D(x) is the determinant of the matrix product such that `D(0)=1, D(-1)=1 and D(2)=7,` then `D(-2)` is equal to

Answer

To solve the problem step by step, we need to find the determinant \( D(x) \) of the product of the matrices and then evaluate \( D(-2) \).

### Step 1: Understand the form of \( D(x) \)
Since the variable \( x \) appears in exactly two places in the determinant, we can conclude that \( D(x) \) is a quadratic polynomial. Therefore, we can express it as:
\[ D(x) = Ax^2 + Bx + C \]

### Step 2: Use the given values to find coefficients
We are given three conditions:
1. \( D(0) = 1 \)
2. \( D(-1) = 1 \)
3. \( D(2) = 7 \)

**Substituting \( D(0) = 1 \):**
\[ D(0) = A(0)^2 + B(0) + C = C = 1 \]
Thus, \( C = 1 \).

**Substituting \( D(-1) = 1 \):**
\[ D(-1) = A(-1)^2 + B(-1) + C = A - B + 1 = 1 \]
This simplifies to:
\[ A - B + 1 = 1 \implies A - B = 0 \implies A = B \]

**Substituting \( D(2) = 7 \):**
\[ D(2) = A(2)^2 + B(2) + C = 4A + 2B + 1 = 7 \]
Since \( A = B \), we can substitute \( B \) with \( A \):
\[ 4A + 2A + 1 = 7 \implies 6A + 1 = 7 \implies 6A = 6 \implies A = 1 \]
Thus, \( B = 1 \) as well.

### Step 3: Write the final form of \( D(x) \)
Now we have:
\[ A = 1, \quad B = 1, \quad C = 1 \]
So, the determinant can be expressed as:
\[ D(x) = x^2 + x + 1 \]

### Step 4: Calculate \( D(-2) \)
Now we need to evaluate \( D(-2) \):
\[ D(-2) = (-2)^2 + (-2) + 1 = 4 - 2 + 1 = 3 \]

### Final Answer
Thus, \( D(-2) \) is equal to:
\[ \boxed{3} \]