Let $T:\mathbb C^3\to\mathbb C^3$ be defined by $T( \left[ {\begin{array}{cc} z_1 \\ z_2 \\ z_3\\ \end{array} } \right] )= \left[ {\begin{array}{cc} z_1-iz_2 \\ iz_1+z_2 \\ z_1+z_2+iz_3\\ \end{array} } \right]$. Then, the adjoint $T^*$ of $T$.
(A)$T^*( \left[ {\begin{array}{cc} z_1 \\ z_2 \\ z_3\\ \end{array} } \right] )= \left[ {\begin{array}{cc} z_1+iz_2 \\ -iz_1+z_2 \\ z_1+z_2-iz_3\\ \end{array} } \right]$
(B)$T^*( \left[ {\begin{array}{cc} z_1 \\ z_2 \\ z_3\\ \end{array} } \right] )= \left[ {\begin{array}{cc} z_1-iz_2+z_3 \\ -iz_1+z_2+z_3 \\ iz_3\\ \end{array} } \right]$
(C)$T^*( \left[ {\begin{array}{cc} z_1 \\ z_2 \\ z_3\\ \end{array} } \right] )= \left[ {\begin{array}{cc} z_1+iz_2+z_3 \\ iz_1+z_2+z_3 \\ iz_3\\ \end{array} } \right]$
(D)$T^*( \left[ {\begin{array}{cc} z_1 \\ z_2 \\ z_3\\ \end{array} } \right] )= \left[ {\begin{array}{cc} iz_1+z_2 \\ z_1-iz_2 \\ z_1-z_2-iz_3\\ \end{array} } \right]$
My attemtpt:- I found matrix with respect to the basis $\{e_1,e_2,e_3\}$. $[T]_{\{e_1,e_2,e_3\}}^{\{e_1,e_2,e_3\}}= \left[ {\begin{array}{cc} 1 & i&1 \\ -i & 1 & 1\\ 0&0&i\\ \end{array} } \right]=A(say) $. Then I found the Hermitian$(A^H)$. Then, I got $T^*( \left[ {\begin{array}{cc} z_1 \\ z_2 \\ z_3\\ \end{array} } \right] )= A^H.\left[ {\begin{array}{cc} z_1 \\ z_2 \\ z_3\\ \end{array} } \right]= \left[ {\begin{array}{cc} z_1+iz_2 \\ -iz_1+z_2 \\ z_1+z_2-iz_3\\ \end{array} } \right]$. (A) is correct according to this steps. Am I correct? Please help me. Is there any simple alternative method to find the adjoint?