If A=begin{bmatrix} sqrt{2} & 1 -1 & sqrt{2} | vedclass

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(B) Given $A=\begin{bmatrix} \sqrt{2} & 1 \ -1 & \sqrt{2} \end{bmatrix}$,we have $\operatorname{det}(A) = (\sqrt{2})(\sqrt{2}) - (1)(-1) = 2 + 1 = 3$

Vedclass · Accepted Answer

$729$

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(B) Given $A=\begin{bmatrix} \sqrt{2} & 1 \ -1 & \sqrt{2} \end{bmatrix}$,we have $\operatorname{det}(A) = (\sqrt{2})(\sqrt{2}) - (1)(-1) = 2 + 1 = 3$.Given $B=\begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix}$,we have $\operatorname{det}(B) = (1)(1) - (0)(1) = 1$.Since $C = ABA^T$,the determinant of $C$ is $\operatorname{det}(C) = \operatorname{det}(A) \cdot \operatorname{det}(B) \cdot \operatorname{det}(A^T)$.Using the property $\operatorname{det}(A^T) = \operatorname{det}(A)$,we get $\operatorname{det}(C) = \operatorname{det}(A)^2 \cdot \operatorname{det}(B) = (3)^2 \cdot 1 = 9$.Now,we need to find $\operatorname{det}(X)$ where $X = A^T C^2 A$.Using the property $\operatorname{det}(MN) = \operatorname{det}(M)\operatorname{det}(N)$,we have $\operatorname{det}(X) = \operatorname{det}(A^T) \cdot \operatorname{det}(C^2) \cdot \operatorname{det}(A)$.Since $\operatorname{det}(C^2) = (\operatorname{det}(C))^2$,we have $\operatorname{det}(X) = \operatorname{det}(A) \cdot (\operatorname{det}(C))^2 \cdot \operatorname{det}(A) = (\operatorname{det}(A))^2 \cdot (\operatorname{det}(C))^2$.Substituting the values,$\operatorname{det}(X) = (3)^2 \cdot (9)^2 = 9 \cdot 81 = 729$.

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

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