If `P=[(1,alpha,3),(1,3,3),(2,4,4)]` is the adjoint of a 3 x 3 matrix A and `|A| = 4`, then `alpha` is equal to

Options

A. 4
B. 11
C. 5
D. 0

Explanation

To solve the problem step by step, we will use the properties of determinants and the relationship between a matrix and its adjoint.

### Step 1: Understand the relationship between the adjoint and the determinant
The determinant of the adjoint of a matrix \( A \) is given by the formula:
\[ |\text{adj}(A)| = |A|^{n-1} \]
where \( n \) is the order of the matrix. For a \( 3 \times 3 \) matrix, \( n = 3 \), so:
\[ |\text{adj}(A)| = |A|^{3-1} = |A|^2 \]

### Step 2: Substitute the given values
We know that \( |A| = 4 \). Therefore:
\[ |\text{adj}(A)| = |A|^2 = 4^2 = 16 \]

### Step 3: Calculate the determinant of matrix \( P \)
Given that \( P = \begin{pmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{pmatrix} \), we need to find \( |P| \) and set it equal to 16. We calculate the determinant of \( P \) using the formula for the determinant of a \( 3 \times 3 \) matrix:
\[ |P| = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \( P \):
- \( a = 1, b = \alpha, c = 3 \)
- \( d = 1, e = 3, f = 3 \)
- \( g = 2, h = 4, i = 4 \)

Thus, we can calculate \( |P| \):
\[ |P| = 1(3 \cdot 4 - 3 \cdot 4) - \alpha(1 \cdot 4 - 3 \cdot 2) + 3(1 \cdot 4 - 3 \cdot 2) \]
\[ = 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6) \]
\[ = 0 - \alpha(-2) + 3(-2) \]
\[ = 2\alpha - 6 \]

### Step 4: Set the determinant equal to 16
Now, we set the determinant equal to 16:
\[ 2\alpha - 6 = 16 \]

### Step 5: Solve for \( \alpha \)
Adding 6 to both sides:
\[ 2\alpha = 22 \]
Dividing by 2:
\[ \alpha = 11 \]

### Conclusion
Thus, the value of \( \alpha \) is:
\[ \alpha = 11 \]