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+18 votes

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. $(AB)'=BA$
asked in Linear Algebra by Veteran (59.8k points)
recategorized by | 1.1k views
Isn't option C and D both are true?
Matrix multiplication is not commutative. So only D.

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

2 Answers

+17 votes
Best answer

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!

answered by Boss (42.7k points)
edited by
What does $A^{'}$ in option (A) means? Is it transpose or inverse?
They didn't say anything about the symbol $"(A)'=(A)^{T}"?$
+14 votes
Answer: D

Given A = A' and B = B'

(AB)' = B'A' = BA
answered by Boss (34k points)

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