Robustapproximatesolutionoflinearequations. We wish to solve the square set of n linear equations Ax=b for the n-vector x. If A is invertible the solution is x=A1b. In this exercise we address an issue that comes up frequently: We don't know A exactly. One simple method is to just choose a typical value of A and use it. Another method, which we explore here, takes into account the variation in the matrix A. We find a set of K versions of A, and denote them as A(1),...,A(k). (These could be found by measuring the matrix A at different times, for example.) Then we choose x so as to minimize A(1)xb2+...+A(K)xb2, the sum of the squares of residuals obtained with the K versions of A.

This choice of x, which we denote xrob, is called a robust (approximate) solution. Give a formula for xrob, in terms of A(1),...,A(K) and b. (You can assume that a matrix you construct has linearly independent columns.) Verify that for K=1 your formula reduces to xrob=(A(1))1b.