If `A=[(1,x,3),(1,3,3),(2,4,4)]` is the adjoint of a `3xx3` matrix B and det. `(B)=4`, then the value of x is ______ .

Explanation

To find the value of \( x \) in the matrix \( A = \begin{pmatrix} 1 & x & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{pmatrix} \), which is the adjoint of a \( 3 \times 3 \) matrix \( B \) with \( \text{det}(B) = 4 \), we can follow these steps:

### Step 1: Use the property of determinants
We know that the determinant of the adjoint of a matrix \( B \) is given by:
\[ \text{det}(\text{adj}(B)) = \text{det}(B)^2 \]
Since \( \text{det}(B) = 4 \), we have:
\[ \text{det}(\text{adj}(B)) = 4^2 = 16 \]

### Step 2: Compute the determinant of matrix \( A \)
Next, we need to calculate the determinant of the matrix \( A \):
\[ A = \begin{pmatrix} 1 & x & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{pmatrix} \]
Using the formula for the determinant of a \( 3 \times 3 \) matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
Where \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \).

Substituting the values:
\[ \text{det}(A) = 1 \cdot (3 \cdot 4 - 3 \cdot 4) - x \cdot (1 \cdot 4 - 3 \cdot 2) + 3 \cdot (1 \cdot 4 - 3 \cdot 2) \]
Calculating each term:
- The first term: \( 1 \cdot (12 - 12) = 0 \)
- The second term: \( -x \cdot (4 - 6) = -x \cdot (-2) = 2x \)
- The third term: \( 3 \cdot (4 - 6) = 3 \cdot (-2) = -6 \)

Putting it all together:
\[ \text{det}(A) = 0 + 2x - 6 = 2x - 6 \]

### Step 3: Set the determinant equal to 16
Since we established that \( \text{det}(A) = 16 \):
\[ 2x - 6 = 16 \]

### Step 4: Solve for \( x \)
Now, we can solve for \( x \):
\[ 2x = 16 + 6 \]
\[ 2x = 22 \]
\[ x = \frac{22}{2} = 11 \]

Thus, the value of \( x \) is \( \boxed{11} \).

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