If A = [√ 2, 1; -1, √ 2], B = [1, 0; 1, 1], C = ABA^T and X = A^T C^2 A, then det X is equal to:

If $\displaystyle \mathrm{A}=\left[\begin{array}{cc}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right], \mathrm{C}=\mathrm{ABA}^{\mathrm{T}}$ and $\displaystyle \mathrm{X}=\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{~A}$, then $\displaystyle \operatorname{det} \mathrm{X}$ is equal to :