To add to the earlier response, and interpreting your question to ask for heuristics for how to manipulate matrices, and not specifically for what matrix multiplication means.
I assume here we interpret vectors as columns vectors, so $x^T$ would refer to a row vector, and capitals for matrices. When $A=(a_{ij})$ then $A^T=(a_{ji})$ so transposition, that is, interchange of rows and columns, corresponds to switching the indices! Remembering that, you can easily convert a symbolic matrix product to a sum over indexed expressions, manipulate, and reconvert to a symbolic matrix product.
One useful trick is pre- and post-multiplication by diagonal matrices: premultiplication corresponds to operations on the rows, while post-multiplication corresponds to operations on the columns. That is, letting $D$ be a diagonal matrix, in $DA$ each row of $A$ is multiplied by the corresponding diagonal element of $D$, while in $AD$ each column of $A$ is multiplied with the corresponding diagonal element.
Now an example to show how to use this manipulative tricks. Suppose $X$ is an $n\times n$ matrix such that there exists an basis for $\mathbb R^n$ consisting of eigenvectors of $X$ (we assume all elements are real here). That is, the eigenvalue/eigenvector equation $Xx=\lambda x$ has $n$ linearly independent solutions, call them (or some choice of them if they are not unique) $x_1, \dots, x_n$. with corresponding eigenvalues $\lambda_i$, the elements of the diagonal matrix $\Lambda$. Write
$$
X x_i = \lambda_i x_i
$$
Now let $P$ be a matrix with the $x_i$ as columns. How can we write the equations above as one matrix equation? Note that the constants $\lambda_i$ are multiplying columns, we know that in the matrix representation the diagonal matrix $\Lambda$ must postmultiply $P$. That is, we get
$$
X P = P \Lambda
$$
Premultiplying on both sides with the inverse of $P$, we get
$$
P^{-1} X P = \Lambda
$$
That is, we can se that $X$ is similar to the diagonal matrix consisting of its eigenvalues.
One more example: If $S$ is a sample covariance matrix, how can we convert it to a sample correlation matrix? The correlation between variable $i$ and $j$ is the covariance divided into the standard deviations of variable $i$ and of variable $j$:
$$
\text{cor}(X_i,X_J) = \frac{\text{cov}(X_i, X_j)}
{\sqrt{\text{var}(X_i) \text{var}(X_j) }}
$$
Looking at this with matrix eyes, we are dividing the $(i,j)$-element of the matrix $S$ with the square roots of the $i$th and $j$th diagonal elements! We are dividing each row of $S$ and each column of $S$ with the same diagonal elements, so it can be expressed as pre- and post-multiplication by the (same) diagonal matrix, that holding the square roots of the diagonal elements of $S$. We have found:
$$
R = D^{-1/2} S D^{-1/2}
$$
where $R$ is the sample correlation matrix, and $D$ is a diagonal matrix holding the diagonal elements of $S$.
There are a lots of applications of this kind of tricks, and I find it so useful that textbooks should include them. One other example: Now let P be a permutation matrix, that is an $n\times n$ matrix representing a permutation on $n$ symbols. Such a matrix has one 1 and $n-1$ zeros in each row and each column, and can be obtained by permuting (in the same way!) the rows and columns of an identity matrix. Now $AP$ (since it is a post-multiplication) permutes the columns of $A$, while $PA$ permutes the rows of $A$.
I don't know how satisfying this answer will be, but I'll give it a shot anyway. The punchline, I think, is that although these "diagonal properties" have just as much aesthetic appeal as their "anti-diagonal" counterparts, the diagonal properties happen to give us information that is more useful for a matrix as it is used mathematically. That is, diagonal symmetry is a more natural thing to look for in the context of linear algebra.
First of all, note that all of these properties are properties of square (that is, $n \times n$) matrices, which are implicitly linear maps from $\Bbb F^n$ to $\Bbb F^n$ (that is, they produce vectors of $n$ entries from vectors of $n$ entries).
The properties that we really care about in linear algebra are the ones that tell us something about how matrices interact with vectors (and ultimately, with other matrices).
Diagonal Matrices
Diagonal matrices are important because they describe a particularly nice class of linear transformations. In particular:
$$
\pmatrix{d_1\\&d_2\\&&\ddots \\ &&& d_n} \pmatrix{x_1\\ x_2\\ \vdots \\ x_n} =
\pmatrix{d_1 x_1\\ d_2 x_2\\ \vdots \\ d_n x_n}
$$
I would say that what a diagonal matrix represents is the fact that each of the $n$ variables required to specify a vector are decoupled. For example, in order to find the new $x_2$, one only needs to look at the old $x_2$, and do what the matrix says.
When we "diagonalize" a matrix, we're finding a way to describe each vector (that is, $n$ independent "pieces of information") that are similarly decoupled as far as the transformation is concerned. So, for example, the matrix
$$
A = \pmatrix{0&1\\4&0}
$$
takes a vector $x = (x_1,x_2)$ and produces a new vector $Ax = (x_2,2x_1)$. There's a nice symmetry to that; in particular, applying $A$ twice gives us the vector $A^2 x = (2x_1,2x_2)$, which is to say that $A$ acts like a diagonal matrix whenever you apply it an even number of times.
However, I would argue that we get a clearer picture of what $A$ does if we diagonalize it. In particular, if we write a vector as $x = a_1(1,2) + a_2(1,-2)$, $A$ gives us the new vector
$$
Ax = a_1 A(1,2) + a_2 A(2,1) = 2a_1(1,2) - 2a_1(1,-2)
$$
In particular, one we know the two pieces of information $a_1$ and $a_2$, we can figure out the new vector using these pieces separately, without having them interact.
So, we see that this antidiagonal $A$ is nice, but just not nearly as simple as the "diagonal version" of the transformation.
Symmetric matrices
Symmetric matrices are particularly nice when we care about dot-products. Dot products are needed whenever you want to think about the angle between vectors in some capacity.
In particular: if we define the dot-product
$$
(x_1,\dots,x_n) \cdot (y_1,\dots,y_n) = x_1y_1 + \cdots x_n y_n
$$
Then a symmetric $A$ will have the property that
$$
(Ax) \cdot y = x \cdot (Ay)
$$
Ultimately, this whole thing connects back to diagonal matrices since every symmetric matrix can be diagonalized in the sense described above. The fact that this can be done is known as the spectral theorem.
Persymmetric matrices, however, don't act in a particularly nice way with respect to any usual operations (like the dot product).
Best Answer
There's not a whole lot you can do to simplify that anymore. In any case, what you've written down is a special case of a symmetric matrix. In general a symmetric $3 \times 3$ matrix will have the form:
$$A= \begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f \end{pmatrix}$$
which has a determinant of $a(df-e^2) + b(ce-bf) + c(be-dc)$. Even worse-looking. The only time it really gets a lot simpler is if you have zeroes in there. The simplest way to calculate is not to calculate.