Explain why solving Ax b is computationally different from

Explain why solving Ax = b is computationally different from doing the product A^-1 b. What is the operation count to obtain A^-1?

Solution

Solving Ax=b directly takes way more operations than doing the product A^-1b

Solving Ax=b takes n(n+1)/2 divisions, (2n^3+3n^2-5n)/6 divisions and (2n^3+3n^2-5n)/6 additions or subtractions.

However computing the product A^-1b takes 2n^3/3 operations to evaluate A^-1,  n(n+1)/2 multiplications and (2n^3+3n^2-5n)/6 additions. Therefore, in all the number of operations are different.

The operation count to obtain A^-1 is 2n^3/3 if A is n*n matrix