. (10 points - Bonus problem) The Fredholm alternative states that: If b ∈ R m is not in the column space of Am×n, then it is not orthogonal to the left nullspace of A. Equivalently, using matrix language, we have that either the system Ax = b OR the system AT y = 0 with y T b = 1 has a solution. Verify the Fredholm alternative for the system Ax = b with A =   1 2 3 4 5 6 7 8 9   and b = (0, 1, 1). (Hint: If Ax = b does not have a solution show that AT y = 0 with y T b = 1 has a solution).