1. Given matrices

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

Also,

$C=ABA^T, \qquad X=A^T C^2 A$

We need to find $\det X$.

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2. Use determinant properties

We use:

• $\det(PQ)=\det P\cdot \det Q$
• $\det(A^T)=\det(A)$
• $\det(C^2)=(\det C)^2$

So,

$\det X=\det(A^T)\det(C^2)\det(A)=\det(A)^2\,\det(C)^2$

Now,

$C=ABA^T$

Hence,

$\det C=\det(A)\det(B)\det(A^T)=\det(A)^2\det(B)$

Therefore,

$\det X=\det(A)^2\left(\det(A)^2\det(B)\right)^2=\det(A)^6\det(B)^2$

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3. Compute $\det A$ and $\det B$

For $A$:

$\det A=(\sqrt{2})(\sqrt{2})-(1)(-1)=2+1=3$

For $B$:

$\det B=(1)(1)-(0)(1)=1$

So,

$\det X=3^6\cdot 1^2=729$

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4. Match with options

$729$

So the correct option is B.

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5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.