For a square matrix B of order 3, if B^T=B^{- | vedclass

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(A) Given that $B$ is an orthogonal matrix, $B^T = B^{-1}$, which implies $B^T B = I$. Taking the determinant on both sides, we get $|B^T B| = |I| = 1

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(A) Given that $B$ is an orthogonal matrix, $B^T = B^{-1}$, which implies $B^T B = I$. Taking the determinant on both sides, we get $|B^T B| = |I| = 1$. Since $|B^T| = |B|$, we have $|B|^2 = 1$. Given $|B| = 1$, we consider the matrix $B-I$. We know that $|B-I| = |B-I|^T = |B^T - I^T| = |B^T - I|$. Since $B^T = B^{-1}$, we have $|B^T - I| = |B^{-1} - I| = |B^{-1}(I - B)| = |B^{-1}| |I - B| = \frac{1}{|B|} |-(B-I)| = \frac{1}{1} (-1)^3 |B-I| = -|B-I|$. Thus, $|B-I| = -|B-I|$, which implies $2|B-I| = 0$, so $|B-I| = 0$.

For a square matrix $B$ of order $3$, if $B^T=B^{-1}$ and $|B|=1$, then $|B-I|=$

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