# Thread: How to prove: A,B are orthogonal matrices, if det(A)+det(B)=0, then A+B is singular.

1. ## How to prove: A,B are orthogonal matrices, if det(A)+det(B)=0, then A+B is singular.

Thanks a lot!!!

2. Originally Posted by seekor
Thanks a lot!!!
Where is the question? seekor, please don't delete posts once you have the answer. Someone else might be able to benefit from them as well.

-Dan

3. Originally Posted by topsquark
The other thread has been locked. The question is:

If $\displaystyle det(A) + det(B) = 0$ then
$\displaystyle det(A + B) = det(A) + det(B) = 0$

Thus the matrix A + B is singular because it has a zero determinant.

-Dan
In general, $\displaystyle det(A + B) \ne det(A) + det(B).$

Example: $\displaystyle A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\ B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix},\ det(A) = det(B) = 0,\ det(A+B) = 1.$

For square matrices, $\displaystyle det(AB) = det(A)det(B).$

4. Originally Posted by JakeD
In general, $\displaystyle det(A + B) \ne det(A) + det(B).$

Example: $\displaystyle A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\ B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix},\ det(A) = det(B) = 0,\ det(A+B) = 1.$

For square matrices, $\displaystyle det(AB) = det(A)det(B).$
(sigh) I thought there was something fishy going on there. I guess I was tired last night! Thanks for the spot.

-Dan

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# A and B are real orthogonal matrices of the some orde4 and detA detB=0 .show that A B is a sungular matrix

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