If `P` is a `3times3` real matrix such that `P^(T)=aP+(a-1)I,` Where `a>I`,then

Options

A. `|Adj|P|gt1`
B. `|Adj|P|=(1)/(2)`
C. `P` is a singular matrix ``
D. `|Adj P| = 1`

Answer

To solve the problem, we start with the equation given:

\[ P^T = aP + (a - 1)I \]

where \( P \) is a \( 3 \times 3 \) real matrix, \( a > 1 \), and \( I \) is the identity matrix.

### Step 1: Take the transpose of both sides

Taking the transpose of both sides gives us:

\[ (P^T)^T = (aP + (a - 1)I)^T \]

Using the property that the transpose of a transpose is the original matrix and the transpose of a sum is the sum of the transposes, we have:

\[ P = aP^T + (a - 1)I \]

### Step 2: Substitute \( P^T \) from the original equation

Now, we can substitute \( P^T \) from the original equation into this equation:

\[ P = a(aP + (a - 1)I) + (a - 1)I \]

### Step 3: Expand and simplify

Expanding the right side:

\[ P = a^2P + a(a - 1)I + (a - 1)I \]

Combining like terms gives:

\[ P = a^2P + (a^2 - a + a - 1)I \]
\[ P = a^2P + (a^2 - 1)I \]

### Step 4: Rearrange the equation

Rearranging the equation, we get:

\[ P - a^2P = (a^2 - 1)I \]

Factoring out \( P \) from the left side:

\[ (1 - a^2)P = (a^2 - 1)I \]

### Step 5: Solve for \( P \)

Now, if we divide both sides by \( (1 - a^2) \) (noting that \( a > 1 \) implies \( 1 - a^2 < 0 \)), we have:

\[ P = \frac{(a^2 - 1)}{(1 - a^2)}I \]

This simplifies to:

\[ P = -I \]

### Step 6: Find the adjoint of \( P \)

The adjoint of a matrix \( P \) is given by:

\[ \text{adj}(P) = \text{det}(P) P^{-1} \]

Since \( P = -I \), we know:

\[ \text{det}(P) = \text{det}(-I) = (-1)^3 \cdot 1 = -1 \]

The inverse of \( P \) is:

\[ P^{-1} = -I \]

Thus, we have:

\[ \text{adj}(P) = (-1)(-I) = I \]

### Step 7: Find the modulus of the adjoint of \( P \)

The modulus of the adjoint of \( P \) is:

\[ |\text{adj}(P)| = |I| = 1 \]

### Conclusion

Thus, the final answer is:

\[ |\text{adj}(P)| = 1 \]