= Solution
Direct <matrix> multiplication gives the multiplication table
$$
J^2=K^2=L^2=-I,\qquad JK=L,\quad KL=J,\quad LJ=K,\qquad KJ=-L,\quad LK=-J,\quad JL=-K.
$$
Together with $I$ acting as the identity, this shows that the product of any two real <linear combinations> is again a real <linear combination>. Thus $\Omega$ is closed under multiplication. It is the <complex matrix representation of quaternions>.
Its general element has the form
$$
\alpha=\begin{pmatrix}a+ib&c+id\\-c+id&a-ib\end{pmatrix}.
$$
If this <matrix> is zero, real and imaginary parts of its first row give $a=b=c=d=0$. Hence $I,J,K,L$ are linearly independent over $\mathbb R$ and form a <basis>:
$$
\boxed{\dim_{\mathbb R}\Omega=4.}
$$
Put $v=bJ+cK+dL$. Since distinct <imaginary units> anticommute, the mixed terms in $v^2$ cancel, leaving $v^2=-(b^2+c^2+d^2)I$. Scalars commute with $v$, so
$$
(aI+v)(aI-v)=a^2I-v^2=(a^2+b^2+c^2+d^2)I.
$$
The reversed product is the same. The coefficient is strictly positive for every nonzero element. Consequently
$$
\boxed{\alpha^{-1}=\frac{aI-bJ-cK-dL}{a^2+b^2+c^2+d^2}\in\Omega\qquad(\alpha\ne0).}
$$
This also identifies $\Omega$ as a real noncommutative <division algebra>, not as the full algebra of complex two-by-two <matrices>.
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