A Matrix Problem

I think statement 3 is incorrect:- For a square matrix $A$ , $\frac{1}{2}(A+A')$ is a symetric matrix not skew symetric matrix:- Consider a square matrix $A$ :- Also let:- $\frac{1}{2}(A+A')=K.........(1)$ Taking transpose on both sides of equation (1):- $\frac{1}{2}(A+A')'=K'......(2)$ Now using the properties well known properties of transpose operator:- $\frac{1}{2}(A+A)'=\frac{1}{2}(A'+(A')')=\frac{1}{2}(A+A').....(3)$ Now putting the value of $\frac{1}{2}(A+A')'$ from equation (3) in (2):- $\frac{1}{2}(A+A')=K'.....(4)$ Clearly LHS of equations (1) and (A) are same hence there RHS must be same so:- $K=K'$ Putting the value of K in above equation:- $\frac{1}{2}(A+A')=\frac{1}{2}(A+A')'$ which is defination of symetric matrix not of skew symetric matrix So i think question may be wrong........it is also possible that i am doing something wrong ,can someone please explain why is statement 3 correct

(1) is incorrect because by definition for symmetric matrix $A = A^T$ that is $a_{ij}$ $=$ $a_{ji}$ for all $i,j$ .

(2) is correct because by definition for skew symmetric matrix $A = - A^T$ that is $a_{ij} = -a_{ji}$ .But now consider the case where $i = j = k$ (say).Here $a_{kk} = - a_{kk}$ implying $a_{kk}$ $=$ $0$ for all $k$ .

for,(3) and (4) see Aman Sharma 's answer it is the perfect way to do it.

P.S. in general any square matrix $A$ $=$ $0.5(A+A^T)$ $+$ $0.5 (A-A^T)$ where $A+A^T$ is symmetric and $A-A^T$ is skew symmetric.

Anuj Shikarkhane, I think all options are incorrect if you don't specify that: $char(K)\neq 2$

How to do by instinct

(1) is wrong as a matrix $(1)$ suffices

then , 3 and 4 can't be both correct, so is is 2 and 4