= Solution
A square <matrix> is invertible precisely when its <determinant> is nonzero. Equivalent conditions are full <rank>, a trivial <kernel>, and zero not being an <eigenvalue>.
For a lower <triangular matrix>, the first row has only the possibly nonzero entry $A_{11}$. A <Laplace expansion> along that row gives $\det A=A_{11}\det A^{(1)}$, where the remaining principal submatrix is again lower triangular. Induction from the one-by-one case yields
$$
\boxed{\det A=\prod_{j=1}^nA_{jj}.}
$$
This proof includes singular triangular matrices; it makes no division by diagonal entries.
For the diagonal example, put $N=n(n+1)/2$. Its <determinant> is $b^{1+2+\cdots+n}=b^N$, and invertibility requires $b\ne0$. Write the <complex number> as $b=re^{i\theta}$ with $r>0$. A nonzero purely imaginary determinant requires $\cos(N\theta)=0$. Thus
$$
\boxed{b=r\exp\left[\frac{(2m+1)\pi i}{n(n+1)}\right],\qquad r>0,\quad m\in\mathbb Z.}
$$
Equivalently $b^N\in i\mathbb R\setminus\{0\}$; the angle is understood modulo $2\pi$.
For <skew-Hermitian matrices>, reversing products under the <conjugate transpose> gives $(CD)^\dagger=D^\dagger C^\dagger=DC$. Consequently
$$
\boxed{CD\text{ is skew-Hermitian}\ \Longleftrightarrow\ DC=-CD\ \Longleftrightarrow\ CD+DC=0.}
$$
The latter condition is vanishing of the <anticommutator>. A universal choice for the last request is
$$
\boxed{\alpha=-1,\qquad CD-DC=[C,D].}
$$
Indeed $(CD-DC)^\dagger=DC-CD=-(CD-DC)$, so the <commutator> of two skew-Hermitian matrices is skew-Hermitian.
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