JEE Mathematics 2023 Question with Solution

Given: $A^T = A$, $B^T = -B$ and $C^T = -C$.

Find: Which of the two statements $S1$ and $S2$ is true.

Let

$$M = A^{13}B^{26} - B^{26}A^{13}$$

Then

$$M^T = (A^{13}B^{26} - B^{26}A^{13})^T$$ $$= (A^{13}B^{26})^T - (B^{26}A^{13})^T$$ $$= (B^T)^{26}(A^T)^{13} - (A^T)^{13}(B^T)^{26}$$

Using $A^T=A$ and $B^T=-B$,

$$M^T = (-B)^{26}A^{13} - A^{13}(-B)^{26}$$

Since $26$ is even,

$$M^T = B^{26}A^{13} - A^{13}B^{26} = -M$$

Hence, $M$ is skew-symmetric. Therefore, statement $S1$ is false.

Now let

$$N = A^{26}C^{13} - C^{13}A^{26}$$

Then

$$N^T = (A^{26}C^{13} - C^{13}A^{26})^T$$ $$= (C^T)^{13}(A^T)^{26} - (A^T)^{26}(C^T)^{13}$$

Using $A^T=A$ and $C^T=-C$,

$$N^T = (-C)^{13}A^{26} - A^{26}(-C)^{13}$$

Since $13$ is odd,

$$N^T = -C^{13}A^{26} + A^{26}C^{13}$$ $$= A^{26}C^{13} - C^{13}A^{26} = N$$

Hence, $N$ is symmetric. Therefore, statement $S2$ is true.

So, only $S2$ is true. The correct option is A.

The solution states "The Correct Option is D", but the working clearly concludes "Only S2 is true," which matches option A in the given options.

For matrix expressions of the form $X-Y$, test symmetry by taking transpose.

• A matrix is symmetric if $M^T=M$.
• A matrix is skew-symmetric if $M^T=-M$.

For $S1$, because $A$ is symmetric, every power of $A$ remains symmetric. Since $B$ is skew-symmetric, $B^{26}$ behaves like an even power, so transpose does not introduce a minus sign overall. Thus the commutator

$$A^{13}B^{26} - B^{26}A^{13}$$

changes to its negative under transpose, so it is skew-symmetric, not symmetric. Hence $S1$ is false.

For $S2$, $A^{26}$ is symmetric and $C^{13}$ carries a negative sign under transpose because the exponent is odd. Therefore,

$$A^{26}C^{13} - C^{13}A^{26}$$

comes back unchanged after transpose, so it is symmetric. Hence $S2$ is true.

Therefore, only $S2$ is true, so the correct option is A.