JEE Main 2024 — 1 February, Shift 1 — Question 3

Compute $\det(A) = (\sqrt{2})(\sqrt{2}) - (1)(-1) = 2 + 1 = 3$. Compute $\det(B) = 1\cdot 1 - 0\cdot 1 = 1$. Since $C = A B A^{T}$, we have $\det(C) = \det(A)\det(B)\det(A^{T}) = (\det(A))^{2}\det(B) = 3^{2}\cdot 1 = 9$. Now $X = A^{T} C^{2} A$, so $\det(X) = \det(A^{T}) \det(C)^{2} \det(A) = (\det(A))^{2} (\det(C))^{2} = 3^{2} \cdot 9^{2} = 9 \cdot 81 = 729$. Thus the answer is 729,

which corresponds to option B.