# 10 Error Bounds for Solving Ax=b     continued...

Note that the original has been entirely ``lost'' from the computation by subtracting from it. We would have gotten the same LU factors whether had been 1, 0, -2, or any number such that . Since the algorithm proceeds to work only with L and U, it will get the same answer for all these different , which correspond to completely different A and so completely different ; there is no way to guarantee an accurate answer. This is called numerical instability, since L and U are not the exact factors of a matrix A+E close to A (another way to say this is that is about as large as ).

Let us see what happens when we go on to solve for x using this LU factorization. The correct answer is . Instead we get the following. Solving yields and ; note that the value 2 has been ``lost'' by subtracting from it. Solving yields and , a completely erroneous solution.

The intuition behind this is that if intermediate quantities in computing the product are very large compared to , then this can lead to loss of accuracy in the factorization of A, because large entries of L and U get added to small entries of A, and these small entries of A get rounded away. If the intermediate quantities in the product were comparable to those of A, however, we would expect a tiny error E=A-LU in the factorization. This is what pivoting tries to guarantee.