731 views

Let $A$ and $B$ be real symmetric matrices of size $n \times n$. Then which one of the following is true?

1. $AA'=I$
2. $A=A^{-1}$
3. $AB=BA$
4. $(AB)'=BA$
recategorized | 731 views
0
Isn't option C and D both are true?
0
Matrix multiplication is not commutative. So only D.
+1

Option $A$ and $B$ are incorrect  because  It is not given in question that matrix $A$ is invertible, it may be the case that $A$ is singular matrix.

Option $C$  is incorrect because Matrix multiplicaton are non commutative

symmetric matrix is a square matrix that is equal to its transpose.

(AB)' = B'A' =BA as Both $A$ and $B$ are symmetric matrices, hence $B' = B$ and $A'=A$

So, (D) is correct option!

Why is $(C)$ not correct option? see the following example:

There are two symmetric matrices given of size $2$x$2$ and $AB != BA$. Therefore (C) is not a correct option!

edited
0
What does $A^{'}$ in option (A) means? Is it transpose or inverse?

Given A = A' and B = B'

(AB)' = B'A' = BA