Proof complexity cramers theorem

Question

I am trying to find a proof for the complexity of using cramers theorem when finding the solution for a Ax=b system. A being a n $\times$ n matrix. It should equal $\mathcal{O}((n+1)!)$

Answer 1

You can compute determinant, characteristic polynomial and the adjunct matrix of $A$ in $O(n^4)$ (almost) without division with the Berkovitz and Leverrier-Faddeev algorithms. These data allow the direct computation of the components of Cramer's formula in additional $O(n^3)$ operations.

Numerical algorithms to compute determinants work in $O(n^3)$ or less. Then again the intermediate results of these algorithms are even closer to the steps of the Gauß algorithm for linear system solving via matrix factorizations.

In the end, except in dimension 2 and 3, perhaps still in 4, there is no useful practical application of Cramer's rule as all attempts to do so efficiently result in standard methods for matrix inversion or linear system solving.

Answer 2

The basic application of Cramer's rule needs the ratio of two $n \times n$ determinants for each variable, with the denominator the same for every variable, so you need $n+1\ (n \times n)$ determinants. If you naively expand a determinant, to get a $n \times n$ determinant takes $n\ ((n-1) \times (n-1))$ determinants. If $D(n)$ is the number of operations to get an $n\times n$ determinant that says $D(n)=nD(n-1)$, which gives $D(n) \sim n!$, so Cramer's rule needs $(n+1)$ operations.