Let `a_r=r^4C_r,b_r=(4-r)^4C_r,A_r=[{:(a_r,2),(3,b_r):}] and A = sum _(r=0)^4A_r` then the value of |A| is equal to

Answer

To solve the problem, we need to find the determinant of a matrix \( A \) defined as follows:

Let:
- \( a_r = r^4 C_r \)
- \( b_r = (4-r)^4 C_r \)

The matrix \( A_r \) is given by:
\[ A_r = \begin{pmatrix} a_r & 2 \\ 3 & b_r \end{pmatrix} \]

We need to compute:
\[ A = \sum_{r=0}^{4} A_r \]

### Step 1: Calculate \( A \)

We can express \( A \) as:
\[ A = \sum_{r=0}^{4} \begin{pmatrix} a_r & 2 \\ 3 & b_r \end{pmatrix} = \begin{pmatrix} \sum_{r=0}^{4} a_r & 2 \sum_{r=0}^{4} 1 \\ 3 \sum_{r=0}^{4} 1 & \sum_{r=0}^{4} b_r \end{pmatrix} \]

### Step 2: Calculate \( \sum_{r=0}^{4} a_r \)

We know that:
\[ a_r = r \cdot 4C_r \]
Thus,
\[ \sum_{r=0}^{4} a_r = \sum_{r=0}^{4} r \cdot 4C_r \]
Using the identity \( r \cdot nC_r = n \cdot (n-1)C_{r-1} \), we can rewrite this as:
\[ \sum_{r=0}^{4} r \cdot 4C_r = 4 \sum_{r=1}^{4} 3C_{r-1} = 4 \cdot 3 \cdot 2^3 = 32 \]

### Step 3: Calculate \( \sum_{r=0}^{4} b_r \)

Similarly, we have:
\[ b_r = (4 - r) \cdot 4C_r \]
Thus,
\[ \sum_{r=0}^{4} b_r = \sum_{r=0}^{4} (4 - r) \cdot 4C_r = 4 \sum_{r=0}^{4} 4C_r - \sum_{r=0}^{4} r \cdot 4C_r \]
Using \( \sum_{r=0}^{4} 4C_r = 2^4 = 16 \):
\[ \sum_{r=0}^{4} b_r = 4 \cdot 16 - 32 = 64 - 32 = 32 \]

### Step 4: Construct the matrix \( A \)

Now substituting these values into the matrix \( A \):
\[ A = \begin{pmatrix} 32 & 10 \\ 15 & 32 \end{pmatrix} \]

### Step 5: Calculate the determinant of \( A \)

The determinant of matrix \( A \) is given by:
\[ |A| = 32 \cdot 32 - 10 \cdot 15 \]
Calculating this:
\[ |A| = 1024 - 150 = 874 \]

### Final Answer

Thus, the value of \( |A| \) is:
\[ \boxed{874} \]