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(B) Given that $A \cdot A^T = A^T \cdot A$ and $B = A^{-1} A^T$. We need to evaluate $B \cdot B^T$. $B \cdot B^T = (A^{-1} A^T) (A^{-1} A^T)^T$. Using

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$B \cdot B^T = I$

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(B) Given that $A \cdot A^T = A^T \cdot A$ and $B = A^{-1} A^T$. We need to evaluate $B \cdot B^T$. $B \cdot B^T = (A^{-1} A^T) (A^{-1} A^T)^T$. Using the property $(XY)^T = Y^T X^T$,we get: $B \cdot B^T = A^{-1} A^T ((A^T)^T (A^{-1})^T)$. Since $(A^T)^T = A$ and $(A^{-1})^T = (A^T)^{-1}$,we have: $B \cdot B^T = A^{-1} A^T A (A^T)^{-1}$. Since $A^T A = A A^T$,we substitute: $B \cdot B^T = A^{-1} (A A^T) (A^T)^{-1}$. Using associative property: $B \cdot B^T = (A^{-1} A) A^T (A^T)^{-1}$. $B \cdot B^T = I \cdot (A^T (A^T)^{-1}) = I \cdot I = I$. Thus,$B \cdot B^T = I$.

If $A$ is a non-singular matrix such that $A \cdot A^T = A^T \cdot A$ and $B = A^{-1} \cdot A^T$,then

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