If `P=[(1,alpha, 3),(1,3,3),(2,4,4)]` is the adjoing of a `3xx3` matrix A and `|A|=4` then `alpha` is equal to

Options

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

Answer

To find the value of \( \alpha \) in the matrix \( P \) given that \( P \) is the adjoint of a \( 3 \times 3 \) matrix \( A \) and \( |A| = 4 \), we can follow these steps:

### Step 1: Write down the matrix \( P \) and the determinant of \( A \)
Given:
\[ P = \begin{pmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{pmatrix} \]
and \( |A| = 4 \).

### Step 2: Use the property of the determinant of the adjoint
The determinant of the adjoint of a matrix \( A \) is given by:
\[ |P| = |A|^{n-1} \]
where \( n \) is the order of the matrix. Here, \( n = 3 \), so:
\[ |P| = |A|^{3-1} = |A|^2 = 4^2 = 16. \]

### Step 3: Calculate the determinant of \( P \)
To find \( |P| \), we can use the determinant formula for a \( 3 \times 3 \) matrix:
\[ |P| = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( P = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \).

For our matrix \( P \):
- \( a = 1, b = \alpha, c = 3 \)
- \( d = 1, e = 3, f = 3 \)
- \( g = 2, h = 4, i = 4 \)

Calculating \( |P| \):
\[ |P| = 1 \cdot (3 \cdot 4 - 3 \cdot 4) - \alpha \cdot (1 \cdot 4 - 3 \cdot 2) + 3 \cdot (1 \cdot 4 - 3 \cdot 2) \]
\[ = 1 \cdot (12 - 12) - \alpha \cdot (4 - 6) + 3 \cdot (4 - 6) \]
\[ = 0 - \alpha \cdot (-2) + 3 \cdot (-2) \]
\[ = 2\alpha - 6. \]

### Step 4: Set the determinant of \( P \) equal to 16
From Step 2, we have:
\[ |P| = 2\alpha - 6 = 16. \]

### Step 5: Solve for \( \alpha \)
Now, we solve the equation:
\[ 2\alpha - 6 = 16. \]
Adding 6 to both sides:
\[ 2\alpha = 22. \]
Dividing by 2:
\[ \alpha = 11. \]

### Conclusion
Thus, the value of \( \alpha \) is \( \boxed{11} \).