If $$A = \begin{pmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{pmatrix}$$, $$B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$$, $$C = ABA^T$$ and $$X = A^TC^2A$$, then $$\det X$$ is equal to:

Given matrices:

$$A = \begin{pmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{pmatrix}$$, $$B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$$.

We need to find $$\det X$$ where $$C = ABA^T$$ and $$X = A^T C^2 A$$.

First, compute $$A^T$$, the transpose of A:

$$A^T = \begin{pmatrix} \sqrt{2} & -1 \\ 1 & \sqrt{2} \end{pmatrix}$$.

Now, compute $$A^T A$$:

$$A^T A = \begin{pmatrix} \sqrt{2} & -1 \\ 1 & \sqrt{2} \end{pmatrix} \begin{pmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{pmatrix} = \begin{pmatrix} (\sqrt{2})(\sqrt{2}) + (-1)(-1) & (\sqrt{2})(1) + (-1)(\sqrt{2}) \\ (1)(\sqrt{2}) + (\sqrt{2})(-1) & (1)(1) + (\sqrt{2})(\sqrt{2}) \end{pmatrix} = \begin{pmatrix} 2 + 1 & \sqrt{2} - \sqrt{2} \\ \sqrt{2} - \sqrt{2} & 1 + 2 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = 3I$$.

Thus, $$A^T A = 3I$$.

Now, $$C = ABA^T$$.

Then, $$C^2 = (ABA^T)(ABA^T) = AB(A^T A)BA^T = AB(3I)BA^T = 3AB^2A^T$$, since scalar multiplication commutes.

Now, compute $$B^2$$:

$$B^2 = B \cdot B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} (1)(1) + (0)(1) & (1)(0) + (0)(1) \\ (1)(1) + (1)(1) & (1)(0) + (1)(1) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$$.

So, $$C^2 = 3A B^2 A^T = 3A \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} A^T$$.

Now, $$X = A^T C^2 A = A^T \left(3A B^2 A^T\right) A = 3 A^T A B^2 A^T A$$.

Substitute $$A^T A = 3I$$:

$$X = 3 (3I) B^2 (3I) = 3 \cdot 3 \cdot 3 \cdot I B^2 I = 27 B^2$$, since the identity matrix commutes.

Thus, $$X = 27 \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 27 & 0 \\ 54 & 27 \end{pmatrix}$$.

Now, compute $$\det X$$:

For a matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$, $$\det = ad - bc$$.

So, $$\det X = (27)(27) - (0)(54) = 729 - 0 = 729$$.

Therefore, $$\det X = 729$$.