Properties of Transpose

Let $$X = A + B$$ and $$Y = A - B$$.

Taking the transpose of $$X$$:

$$X^T = (A + B)^T = A^T + B^T = A - B = Y$$

Taking the transpose of $$Y$$:

$$Y^T = (A - B)^T = A^T - B^T = A - (-B) = A + B = X$$

Therefore, $$X^T = Y$$ and $$Y^T = X$$.

2. Analyzing Statement S1 ($$P$$)

The matrix $$P$$ is defined as $$P = Y^{2023}X^{2023} - X^{2023}Y^{2023}$$.

To check if it is skew-symmetric, find $$P^T$$:

$$P^T = (Y^{2023}X^{2023})^T - (X^{2023}Y^{2023})^T$$

Using the property $$(MN)^T = N^T M^T$$:

$$P^T = (X^T)^{2023}(Y^T)^{2023} - (Y^T)^{2023}(X^T)^{2023}$$

Substitute $$X^T = Y$$ and $$Y^T = X$$:

$$P^T = Y^{2023}X^{2023} - X^{2023}Y^{2023}$$ $$P^T = P$$

Since $$P^T = P$$, matrix $$P$$ is symmetric, not skew-symmetric.

Result: S1 is False.

3. Analyzing Statement S2 ($$Q$$)

Substitute $$A^4 + B^4 = A$$ into the expression for $$Q$$:

$$Q = (A + B)^T A (A - B) + (A - B)^T A (A + B)$$

Using the transpose relationships $$X^T = Y$$ and $$Y^T = X$$:

$$Q = Y A Y + X A X$$

Now, find $$Q^T$$:

$$Q^T = (Y A Y)^T + (X A X)^T$$ $$Q^T = Y^T A^T Y^T + X^T A^T X^T$$

Substitute $$Y^T = X$$, $$X^T = Y$$, and $$A^T = A$$:

$$Q^T = X A X + Y A Y$$ $$Q^T = Q$$

Since $$Q^T = Q$$, matrix $$Q$$ is symmetric.

Result: S2 is True.

Conclusion

• S1 is False (P is symmetric).
• S2 is True (Q is symmetric).

Correct Option: C (Only S2 is true)

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

$$P^T = 2P + I \quad \text{--- (Equation 1)}$$
Take the Transpose of Both Sides:
$$(P^T)^T = (2P + I)^T$$
Since $$(P^T)^T = P$$ and $$(A+B)^T = A^T + B^T$$:
$$P = 2P^T + I^T$$
Since the transpose of the identity matrix is itself ($$I^T = I$$):
$$P = 2P^T + I \quad \text{--- (Equation 2)}$$
 Replace the $$P^T$$ in Equation 2 with $$(2P + I)$$:
$$P = 2(2P + I) + I$$
$$P = 4P + 2I + I$$
$$P = 4P + 3I$$
$$P = -I$$
: We are given the eigenvalue equation $$PX = \lambda X$$. Substitute $$P = -I$$:
$$(-I)X = \lambda X$$
$$-X = \lambda X$$
Since $$X \neq 0$$, comparing both sides directly yields:
$$\lambda = -1$$