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(A) Since $A$ and $B$ are orthogonal matrices,we have $AA^{	op} = I$ and $BB^{	op} = I$. Taking the determinant on both sides,we get $\operatorname{

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$A+B$ is singular

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(A) Since $A$ and $B$ are orthogonal matrices,we have $AA^{	op} = I$ and $BB^{	op} = I$. Taking the determinant on both sides,we get $\operatorname{det}(A)\operatorname{det}(A^{	op}) = 1$ and $\operatorname{det}(B)\operatorname{det}(B^{	op}) = 1$. Since $\operatorname{det}(A^{	op}) = \operatorname{det}(A)$,we have $(\operatorname{det}(A))^2 = 1$ and $(\operatorname{det}(B))^2 = 1$,which implies $\operatorname{det}(A) = \pm 1$ and $\operatorname{det}(B) = \pm 1$. Given $\operatorname{det}(A) + \operatorname{det}(B) = 0$,we have $\operatorname{det}(A) = -\operatorname{det}(B)$. Now,consider $\operatorname{det}(A+B) = \operatorname{det}(A(I + A^{-1}B))$. Since $A$ is orthogonal,$A^{-1} = A^{	op}$. Thus,$\operatorname{det}(A+B) = \operatorname{det}(A(I + A^{	op}B))$. We can write $A+B = A(I + A^{	op}B) = A(B^{	op}B + A^{	op}B) = A(B^{	op} + A^{	op})B$. Taking the determinant: $\operatorname{det}(A+B) = \operatorname{det}(A) \operatorname{det}(B^{	op} + A^{	op}) \operatorname{det}(B) = \operatorname{det}(A)\operatorname{det}(B) \operatorname{det}((A+B)^{	op})$. Since $\operatorname{det}(A) = -\operatorname{det}(B)$,we have $\operatorname{det}(A)\operatorname{det}(B) = -(\operatorname{det}(B))^2 = -1$. Thus,$\operatorname{det}(A+B) = -1 \cdot \operatorname{det}(A+B)$. This implies $2 \operatorname{det}(A+B) = 0$,so $\operatorname{det}(A+B) = 0$. Therefore,$A+B$ is a singular matrix.

Let $A$ and $B$ be orthogonal matrices and $\operatorname{det}(A) + \operatorname{det}(B) = 0$. Then

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