JEE Main 2024 — 29 January, Shift 1 — Question 19

Given $AA^T = I$, it follows that $A^T = A^{-1}$ and $A^T A = I$. Expand $(A+A^T)^2 = A^2 + AA^T + A^T A + (A^T)^2 = A^2 + I + I + (A^T)^2 = A^2 + 2I + (A^T)^2$. Expand $(A-A^T)^2 = A^2 - AA^T - A^T A + (A^T)^2 = A^2 - I - I + (A^T)^2 = A^2 - 2I + (A^T)^2$. Add the two results: $(A+A^T)^2+(A-A^T)^2 = 2A^2 + 2(A^T)^2$. Multiply by $A$: $A\left[(A+A^T)^2+(A-A^T)^2\right] = 2A^3 + 2A(A^T)^2$. Since $A^T = A^{-1}$, we have $(A^T)^2 = A^{-2}$, so $A(A^T)^2 = A A^{-2} = A^{-1} = A^T$. Thus $A\left[(A+A^T)^2+(A-A^T)^2\right] = 2A^3 + 2A^T$. Multiplying by the correct factor $\frac12$ gives $\frac12 \times (2A^3 + 2A^T) = A^3 + A^T$,

corresponding to option D.