Let `A=[(3x^2),(1),(6x)],B=[(a,b,c)],and C=[((x+2)^2,5x^2,2x),(5x^2,2x,(x+2)^2),(2x,(x+2)^2,5x^2)]` be three given matrices, where `a ,b ,c`` ,"and" ``x in R dot` Given that `f(x)=a x^2+b x+c ,` then the value of `f(I)` is ______.

Explanation

To solve the problem step by step, we will follow the procedure outlined in the video transcript and derive the necessary values systematically.

### Step-by-Step Solution:

1. **Identify the Matrices**:
- Let \( A = \begin{pmatrix} 3x^2 & 1 & 6x \end{pmatrix} \)
- Let \( B = \begin{pmatrix} a & b & c \end{pmatrix} \)
- Let \( C = \begin{pmatrix} (x+2)^2 & 5x^2 & 2x \\ 5x^2 & 2x & (x+2)^2 \\ 2x & (x+2)^2 & 5x^2 \end{pmatrix} \)

2. **Matrix Multiplication**:
- We need to compute the product \( AB \):
\[ AB = A \cdot B = \begin{pmatrix} 3x^2 & 1 & 6x \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = 3ax^2 + b + 6cx \]

3. **Trace of Matrix C**:
- The trace of matrix \( C \) is the sum of its diagonal elements:
\[ \text{Trace}(C) = (x+2)^2 + 2x + 5x^2 \]

4. **Equate the Traces**:
- From the problem statement, we have:
\[ \text{Trace}(AB) = \text{Trace}(C) \]
Thus,
\[ 3ax^2 + 6cx + b = (x+2)^2 + 2x + 5x^2 \]

5. **Expand the Right Side**:
- Expanding \( (x+2)^2 \):
\[ (x+2)^2 = x^2 + 4x + 4 \]
- Therefore,
\[ \text{Trace}(C) = x^2 + 4x + 4 + 2x + 5x^2 = 6x^2 + 6x + 4 \]

6. **Compare Coefficients**:
- Now we equate coefficients from both sides:
- For \( x^2 \): \( 3a = 6 \) → \( a = 2 \)
- For \( x \): \( 6c = 6 \) → \( c = 1 \)
- For the constant term: \( b = 4 \)

7. **Form the Function \( f(x) \)**:
- The function is given by:
\[ f(x) = ax^2 + bx + c = 2x^2 + 4x + 1 \]

8. **Calculate \( f(1) \)**:
- Substitute \( x = 1 \):
\[ f(1) = 2(1)^2 + 4(1) + 1 = 2 + 4 + 1 = 7 \]

### Final Answer:
The value of \( f(1) \) is \( 7 \).

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