Determinant of the Adjoint

For a $$3\times3$$ matrix,

$$\det(\operatorname{adj}A)=(\det A)^2$$

Given,

$$\det(A)=2$$

After applying adjoint repeatedly,

$$n=2^{2^{2025}}$$

Now powers of $$2$$ modulo $$9$$ repeat with period $$6$$:

$$2^1\equiv2, \qquad2^2\equiv4,\qquad 2^3\equiv8,\qquad 2^4\equiv7,\qquad 2^5\equiv5,\qquad 2^6\equiv1\pmod9$$

Hence,

$$2^{2^{2025}} \equiv2^{\,2^{2025}\bmod6}\pmod9$$

Also,

$$2^n\equiv4\pmod6 \quad\text{for all }n\ge2$$

Therefore,

$$2^{2025}\equiv4\pmod6$$

Hence,

$$2^{2^{2025}} \equiv2^4\pmod9$$

$$\equiv16\pmod9$$

$$\equiv7\pmod9$$

Therefore, the required remainder is

$$\boxed7$$

Given the equation:

$$P^T = kP + (k^2 - 1)I \quad \dots \text{(1)}$$

Taking the transpose on both sides of equation (1):

$$(P^T)^T = (kP + (k^2 - 1)I)^T$$ $$P = kP^T + (k^2 - 1)I \quad \dots \text{(2)}$$

Substitute the value of $$P^T$$ from equation (1) into equation (2):

$$P = k[kP + (k^2 - 1)I] + (k^2 - 1)I$$ $$P = k^2 P + k(k^2 - 1)I + (k^2 - 1)I$$

Rearranging the terms to isolate $$P$$:

$$P - k^2 P = [k(k^2 - 1) + (k^2 - 1)]I$$ $$P(1 - k^2) = (k^2 - 1)(k + 1)I$$

To solve for $$P$$, we can rewrite $$(1 - k^2)$$ as $$-(k^2 - 1)$$:

$$-P(k^2 - 1) = (k^2 - 1)(k + 1)I$$

Dividing by $$(k^2 - 1)$$ (as $$k \neq 1$$ and $$P$$ is non-singular):

$$-P = (k + 1)I \implies P = -(k + 1)I$$

Since $$P$$ is a $$3 \times 3$$ matrix, we use the property $$|cA| = c^n |A|$$:

$$|P| = |-(k + 1)I|$$ $$|P| = [-(k + 1)]^3 |I|$$ $$|P| = -(k + 1)^3$$

Using the determinant property of adjoints for an $$n \times n$$ matrix, $$|\text{adj}(P)| = |P|^{n-1}$$. For $$n = 3$$:

$$|\text{adj}(P)| = |P|^2$$ $$|\text{adj}(P)| = [-(k + 1)^3]^2$$ $$|\text{adj}(P)| = (k + 1)^6$$

Since $$(k + 1)^6$$ is the same as $$(1 + k)^6$$, we reach our final result.

Correct Option: A

To solve this problem, we need to apply three key properties of determinants:

1. Scalar Multiplication Property: For any $$n \times n$$ matrix $$M$$ and a scalar $$k$$:$$\det(kM) = k^n \det(M)$$
2. Product Property: For any square matrices $$M$$ and $$N$$ of the same size:$$\det(MN) = \det(M) \cdot \det(N)$$
3. Adjoint Property: For any $$n \times n$$ matrix $$M$$:$$\det(\text{adj}(M)) = (\det(M))^{n-1}$$

Step 1: Set up the determinant expression

We are given the equation:

$$B = 2A \cdot \text{adj}(A)$$

Taking the determinant on both sides:

$$\det(B) = \det(2A \cdot \text{adj}(A))$$

Step 2: Extract the scalar factor

The matrix inside is a $$4 \times 4$$ matrix (since $$A$$ is $$4 \times 4$$, its adjoint is also $$4 \times 4$$, and their product is $$4 \times 4$$).

Using the property $$\det(kM) = k^n \det(M)$$ with scalar $$k = 2$$ and dimension $$n = 4$$:

$$\det(B) = 2^4 \cdot \det(A \cdot \text{adj}(A))$$ $$\det(B) = 16 \cdot \det(A \cdot \text{adj}(A))$$

Step 3: Split the matrix product

Using the product property $$\det(MN) = \det(M) \cdot \det(N)$$:

$$\det(B) = 16 \cdot \det(A) \cdot \det(\text{adj}(A))$$

Step 4: Substitute the given values

We know that $$\det(A) = 3$$. Using the adjoint identity $$\det(\text{adj}(A)) = (\det(A))^{n-1}$$ where $n = 4$:

$$\det(\text{adj}(A)) = 3^{4-1} = 3^3 = 27$$

Now, substitute $$\det(A) = 3$$ and $$\det(\text{adj}(A)) = 27$$ back into our equation:

$$\det(B) = 16 \cdot 3 \cdot 27$$ $$\det(B) = 48 \cdot 27$$ $$\det(B) = 1296$$

This perfectly matches Option B.