Math 1b Practical — Problem Set 2Due 3:00pm, Tuesday, January 18, 20111. (15 points) Even when a system Ax = b of linear equations has no solutions, we may askfor “approximate solutions”. If A and b are real, it can be shown that AAx = Ab alwayshas solutions x, and the solutions of this latter system are called least-squares approximatesolutions of Ax = b. We will see later in the term why this name is appropriate and inwhat sense they are good approximate solutions.LetA =⎛⎜⎝11121314⎞⎟⎠and b =⎛⎜⎝571012⎞⎟⎠.In this case there are no (exact) solutions to Ax = b, but find the least-squares approximatesolution x =[s, t](there will be only one, in this case). Calculate Ax.Howclosetob isit (give the Eucliean distance in R4)?2. (20 points) Let A be a matrix and b a column vector, of the same height.(i) Prove that it is not possible to find both (a) a column vector x such that Ax = b,and (b) a row vector y such that yA = 0 but yb =0.(ii) Explain how, in the case that Ax = b has no solutions, to find a row vector y suchthat yA = 0 but yb = 0. It is not necesssary to prove anything. Just explain carefullyhow to do this (describe an algorithm) and illustrate with the exampleA =⎛⎝123234345⎞⎠, b =⎛⎝013⎞⎠.3. (25 points) The following two matrices represent nonhomogeneous systems of four linearequations Ax = b in seven unknowns (the last column contains the vector b of constantterms). For each matrix find either (1) a basic solution (x1,...,x7) in nonnegative numbers,or (2) prove that there is no nonnegative solution by producing a row vector y with yA ≥ 0but yb < 0.⎛⎜⎝59105 1 53503 4 5101061954523526387017⎞⎟⎠⎛⎜⎝1 5 3 5 5 9 10 10507 3 54 3 78811054 6 6479 8 4910 1⎞⎟⎠You will probably want some software to do the pivots for you. Help for Mathematicawill be posted later.4. (20 points) Vectors v1, v2,...,vnare affinely dependent provided thatc1v1+ c2v2+ ...+ cnvn= 0for some scalars c1,c2,...,cn, not all zero, with c1+ c2+ ...+ cn= 0. For example, in R2,the vectors (2, −1), (3, 2), and (4, 5) are affinely dependent, since1(2, −1) − 2(3, 2) + 1(4, 5) = (0, 0)and the sum of the coefficients is 1 − 2 + 1 = 0. (Geometrically, three points in R2areaffinely dependent if and only if they lie on a line—more about this later.)(i) Prove that v1, v2,...,vnare affinely dependent if and only ifv2− v1, v3− v1,...,vn− v1are linearly dependent.(ii) Suppose v1,...,vn∈ Rm.Provethatv1, v2,...,vnare affinely dependent if andonly if the vectors(v1, 1), (v2, 1),...,(vn, 1) ∈ Rm+1are linearly dependent. Here (v, 1) denotes the result of appending a 1 as an (m +1)-stcoordinate to the m coordinates of v.5. (20 points) Let U be the span of the three vectors(2, 1, −1, 3), (1, 0, 2, −1), (5, 1, 5, 0).(i) Find a basis for U . What is the dimension of U ? Show some work.(ii) Find a basis for U⊥. What is the dimension of U⊥?(iii) Find a homogeneous system of linear equations whose solution space is