Determinants of Complex Matrices[MAT204 C.A solution]

(a)Find the Determinant of $$\begin{equation} A=\left[\begin{array}{ccc} 0 & 1+i & 1+2i\\ 1-i & 0 & 2-3i\\1-2i & 2+3i & 0\end{array}\right] \end{equation}$$
(b)Hence verify whether or not $det(A)=det(a^T)$
(c)show that $\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)=A$

SOLUTION: firstly we compute the determinant of A
i.e $$\begin{equation}det(A)=0[(0\times0)-(2-3i)(2+3i)]-(1+i)[(1-i)(0)-(2-3i)(1-2i)]\end{equation}$$

$$\begin{equation} +(1+2i)[(1-i)(2+3i)-(0)(1-2i)] \end{equation}$$
$$\begin{equation} =3-11i+3+11i=6 \end{equation}$$
(b) now we want to verify whether $det(A)=det(a^T)$ since we
already know det(A)=6,it remains to show the L.H.S
$$\begin{equation} A^T=\left[\begin{array}{ccc} 0 & 1-i & 1-2i\\ 1+i & 0 & 2+3i\\1+2i & 2-3i & 0\end{array}\right] \end{equation}$$
$$\begin{equation} det(A^T)=0-(1-i)[(1+i)(0)-(2+3i)(1+2i)]+(1-2i)[(1+i)(2-3i)-0(1+2i)] \end{equation}$$
$$\begin{equation} =3+11i+3-11i=6 \end{equation}$$
hence $det(A)=det(A^T)$
(c)show that $\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)=A$ this
can simply be evaluated by following solving each matrix step by step.
we know $$\begin{equation} A=\left[\begin{array}{ccc} 0 & 1+i & 1+2i\\ 1-i & 0 & 2-3i\\1-2i & 2+3i & 0\end{array}\right] \end{equation}$$ and $$\begin{equation} A^T=\left[\begin{array}{ccc} 0 & 1-i & 1-2i\\ 1+i & 0 & 2+3i\\1+2i & 2-3i & 0\end{array}\right] \end{equation}$$
now we first compute for $$\begin{equation}A+A^T=\left[\begin{array}{ccc} 0 & 1+i & 1+2i\\ 1-i & 0 & 2-3i\\1-2i & 2+3i & 0\end{array}\right] + \left[\begin{array}{ccc} 0 & 1-i & 1-2i\\ 1+i & 0 & 2+3i\\1+2i & 2-3i & 0\end{array}\right] \end{equation}$$
$$\begin{equation}= \left[\begin{array}{ccc} 0 & 2 & 2\\ 2 & 0 & 4\\2 & 4 & 0\end{array}\right] \end{equation}$$
now it remains for $A-A^T$
$$\begin{equation} A-A^T=\left[\begin{array}{ccc} 0 & 1+i & 1+2i\\ 1-i & 0 & 2-3i\\1-2i & 2+3i & 0\end{array}\right] - \left[\begin{array}{ccc} 0 & 1-i & 1-2i\\ 1+i & 0 & 2+3i\\1+2i & 2-3i & 0\end{array}\right]\end{equation}$$
$$\begin{equation}= \left[\begin{array}{ccc} 0 & 2i & 4i\\ 2i & 0 & -6i\\-4i & 6i & 0\end{array}\right] \end{equation}$$
we now proceed to showing for $$\begin{equation}\frac{1}{2}(A+A^T)=\frac{1}{2}\left[\begin{array}{ccc} 0 & 2 & 2\\ 2 & 0 & 4\\2 & 4 & 0\end{array}\right]\end{equation}$$
$$\begin{equation} =\left[\begin{array}{ccc} 0 & 1 & 1\\ 1 & 0 & 2\\1 & 2 & 0\end{array}\right] \end{equation}$$
and $$\begin{equation} \frac{1}{2}(A-A^T)=\frac{1}{2}\left[\begin{array}{ccc} 0 & 2i & 4i\\ 2i & 0 & -6i\\-4i & 6i & 0\end{array}\right] \end{equation}$$
$$\begin{equation} =\left[\begin{array}{ccc} 0 & i & 2i\\ i & 0 & -3i\\-2i & 3i & 0\end{array}\right] \end{equation}$$
adding $\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)$
$$\begin{equation} =\left[\begin{array}{ccc} 0 & 1 & 1\\ 1 & 0 & 2\\1 & 2 & 0\end{array}\right] \left[\begin{array}{ccc} 0 & i & 2i\\ i & 0 & -3i\\-2i & 3i & 0\end{array}\right] \end{equation}$$
$$\begin{equation} \left[\begin{array}{ccc} 0 & 1+i & 1+2i\\ 1-i & 0 & 2-3i\\1-2i & 2+3i & 0\end{array}\right] \end{equation}$$
and hence $\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)=A$