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
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