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

• Sep 4th 2007, 01:30 PM
seekor
How to prove: A,B are orthogonal matrices, if det(A)+det(B)=0, then A+B is singular.
Thanks a lot!!!
• Sep 4th 2007, 02:57 PM
topsquark
Quote:

Originally Posted by seekor
Thanks a lot!!!

:mad: 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
• Sep 4th 2007, 06:06 PM
JakeD
Quote:

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, $\displaystyle det(AB) = det(A)det(B).$
• Sep 5th 2007, 06:43 AM
topsquark
Quote:

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