Answer: Option C. -> A(z)=A(x)A(y)
:
C
$A\left(z\right)=A\left(\frac{x+y}{1+xy}\right)=\left[\frac{1+xy}{(1-x)(1-y)}\right]\left[\begin{array}{cc}1& -\left(\frac{x+y}{1+xy}\right)\\ -\left(\frac{x+y}{1+xy}\right)& 1\end{array}\right]$
$\therefore A\left(x\right).A\left(y\right)=A\left(z\right)$

Answer: Option B. -> symmetric
:
B
We are given that the matrix A is skew symmetric which implies that,
$A^T=-A$ --------(1)
$(A^4{)}^T=(A.A.A.A.{)}^T=A^T{A}^T{A}^T{A}^T$ (By applying the property of transpose of a matrix)
$\Rightarrow$ (-A) (-A) (-A) (-A) (Using equation (1))
$=(-1{)}^4{A}^4=A^4$
i.e,$(A^4{)}^T=(A^4)$
$A^4$ is symmetric.

Answer: Option C. -> A - I is non zero singular
:
C
$A^2=I\Rightarrow A^2-I=0$
$\Rightarrow$ (A+I)(A-I)=0
$\therefore$ either $|A+I|=0$ or
$|A-I|=0$
If $|A-I|\ne 0$, then $(A+I)(A-I)=0\Rightarrow A+I=0$ which is not so
$\therefore |A-I|$ and $A-I\ne 0$.

Answer: Option C. -> Re(aij)
:
C
By taking complex conjugate of a matrix we reverse the sign of imaginary parts of all the elements in the original matrix. i.e., if the element in A is x + iy, then the corresponding element in $A^{\ast }$is x - iy.
So when A and $A^{\ast }$is added the imaginary parts cancel out and the sum becomes 2 times the real part of element in A.
i.e., since $\left(a_{ij}\right)$ is general element in A, the general element in $A+A^{\ast }$becomes $2Re\left(a_{ij}\right).$
$\therefore$General element in $\frac{A+A^{\ast }}{2}=Re\left(a_{ij}\right).$