Question: Robust approximate solution of linear equations. We wish to solve the square set of \(n\) linear equations \(A x=b\) for the \(n\) -vector \(x\). If \(A\) is invertible the solution is \(x=A^{-1} b\). 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 \(A^{(1)}, \ldots, A^{(K)}\). (These could be found by measuring the matrix \(A\) at different times, for example.) Then we choose \(x\) so as to minimize

the sum of the squares of residuals obtained with the \(K\) versions of \(A\). This choice of \(x\), which we denote \(x^{\text {rob }}\), is called a robust (approximate) solution. Give a formula for \(x^{\text {rob }}\), in terms of \(A^{(1)}, \ldots, A^{(K)}\) and \(b\). (You can assume that a matrix you construct has linearly independent columns.) Verify that for \(K=1\) your formula reduces to \(x^{\mathrm{rob}}=\left(A^{(1)}\right)^{-1} b\)

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