Well, one trivial connection is that if you look at $1\times 1$ matrices (which have only a single complex entry), then you'll find that it is real iff it is Hermitian, its complex conjugate is its conjugate transpose, and its polar decomposition is the polar form.
Also, just like a complex number can be uniquely decomposed into a real and an imaginary part ($z = a+\mathrm ib$ with real $a,b$), a complex matrix can be uniquely decomposed into a Hermitian and an "anti-Hermitian" part, i.e,. $M =A + \mathrm iB$ with $A$ and $B$ Hermitian. And just like $\Re(z)=\frac12(z+\bar z)$ and $\Im(z)=\frac1{2\mathrm i}(z-\bar z)$, the Hermitian part of a matrix is $\frac12(M+M^*)$ and the "anti-Hermitian" part is $\frac1{2\mathrm i}(M-M^*)$.
Moreover, just like $\bar zz$ is a non-negative real number, $M^*M$ is a positive semidefinite matrix.
Another point: Hermitian matrices have real eigenvalues, and unitary matrices have eigenvalues of the form $\mathrm e^{\mathrm i\phi}$.
About the usefulness of the analogy:
In classical physics, observables should be real. In quantum physics, observables are represented by Hermitian matrices. Also, the quantum analogue to probability densities, which are non-negative functions with integral $1$, are density operators, which are positive semidefinite matrices with trace $1$. So there's indeed some connection.
Let $A$ be an arbitrary matrix, and suppose that it we can reduce $A$ to $R$. Then there is an invertible matrix $M$ such that $A = MR$.
The only thing that matrices $A$ and $R$ necessarily have in common is that they have the same nullspace. Outside of that (and any consequences), $A$ and $R$ share no notable properties.
In particular, $A$ and $R$ do not generally have the same eigenvalues. If that were the case, we could argue that any invertible matrix is reducible to the identity matrix and therefore must have the eigenvalue $1$.
Best Answer
The conjugate of a real number is just the number itself. So the adjoint of a real matrix is its transpose. For the example you give, if $$A = \begin{bmatrix}1 & 1 \\ 1 & -1 \\ 0 & 1\end{bmatrix}$$ then $$A^* = \begin{bmatrix}1 & 1 & 0\\ 1 & -1 & 1\end{bmatrix}.$$