Let m x n real matrix A have rank n. Then although for given b = Rm the linear system Ax = b may not have a true solution x R, the normal equations AT Ax = ATb (1) (2) always have a unique solution x", say. We call the x the least squares solution of (1). In particular, since ATA is nonsingular, as Rank(A)=n, then from (2) we have x=(ATA)- Ab In practice, n and m are very large, and so one does not explicitly compute (ATA). Instead: let A = UVT (3) be the singular value decomposition (SVD) of A. That is, U is an m x m orthogonal matrix, V is an n x n orthogonal matrix, and m x n matrix is of the form 01 02 (4) where 1 2022 2020, and the off-diagonal entries of are zero. Therewith, do the following: (a) Show that the least squares solution of (1) has the expression x=V(ETC)-TUTb. (5) where U,E, and V are the matrices from the SVD (3). Moreover, write down the explicit form of (TE)-1 (b) Use the expression (5) to find the least squares solution of the following problem: 21+2+3 = 4 -1+2+3 = 0 -22+231 2123= 2. You must use the expression (5) for the possibility of credit.