Why Gaussian method is recommended for $4\times4$ determinant? I wanted to know why Gaussian elimination method in Linear Algebra has order of $n^3$
for calculation and found this.But I don't know why this:

From the table of $n^3$ vs $n!$ we see that $4^3 > n!$, yet the Gaussian method is the one that is
  generally recommended for calculating $4 × 4$ determinants, why?

Also, if you can suggest other link for "why Gaussian elimination method in Linear Algebra has order of $n^3$", I will be obliged. 
 A: When evaluating algorithms such as Gaussian elimination and cofactor expansion, the primary operations are addition and multiplication.  Of these, multiplication is usually thought to be much slower than addition, so it makes sense to just count the multiplications (and divisions).
It is not hard to show that using Gaussian elimination on an $n\times n$ matrix to compute the determinant typically requires $(n^3+2n-3)/3$ multiplications.
The Leibniz formula for determinants has $n!$ terms, each with $n-1$ multiplications, and therefore has $(n-1)n!$ multiplications.
The method of Laplace cofactor expansion is somewhat faster, since it collects terms together before mutliplying.  It only takes $n(2^{n-1}-1)$ multiplications.
Using these formulas, we can make a table showing the difficulty of each method:
$$
\begin{array}{|c|c|c|c|}
\hline
n & 2 & 3 & 4 & 5 & 6 \\
\hline
\mathrm{Gauss} & 3 & 10 & 23 & 44 & 75 \\
\hline
\mathrm{Leibniz} & 2 & 12 & 72 & 480 & 3600\\
\hline
\mathrm{Laplace} & 2 & 9 & 28 & 75 & 186 \\
\hline
\end{array}
$$
As you can see, the Laplace expansion is best for $2\times 2$ and $3\times 3$, but Gaussian elimination becomes better starting with $4\times 4$ matrices.  (Though it's very close -- computing a $4\times 4$ determinant using cofactor expansion only takes 22% more time.)
Note that the formulas above are exact.  When you read that Gaussian elimination has running time "on the order of $n^3$", all this means is that the formula for the number of steps is some kind of cubic polynomial.  All cubic polynomials grow at roughly the same rate, up to multiplication by a constant, so saying that Gaussian elimination has running time on the order of $n^3$ tells you approximately how quickly the running time grows.  But it makes no sense to plug specific numbers into $n^3$, because $n^3$ is only an approximate formula -- specifically, it's an approximation for $(n^3+2n-3)/3$.
