2. A linear system with no free variables: Consider the linear system x1 + 2x2 + 3x3 2x1 - x2 + 3x3 4x1 - x2 + 12x3 You'll solve this linear system in a few ways: = 115 = 1421 = 4214 (a) First, enter the coefficient matrix for this system as the variable "A", and the right- hand side as the column vector "b". Next, enter the command "A \b", which outputs the (in this case, unique) solution to this linear system. (b) (*) Second, enter the augmented matrix as the variable "M". In the next line, run the command rref (M), which outputs the row reduced echelon form of M. Note that there are no free variables. Explain, in words, why in this case we have that the last column of M is the (unique) solution to the original linear system. If you did everything right, you should find that this solution is the same as that you found in part (a). 4. A linear system with a free variable: Using only what we have so far, MATLAB does not provide parametric solutions to an underdetermined system of equations. To see this, consider the linear system consisting of the first two equations in the linear system in the previous Problem 2. Enter the backslash command A \ b (for appropriate A, b) to a single solution. On the other hand, we know from class that, if such a system is consistent, there must be infinitely many solutions: provide an explanation as to why this is true. At best, using the tools at our disposal, we will have to read the set of solutions off of the RREF. For this, enter the appropriate augmented matrix and take its RREF using MATLAB. Based on this, identify vectors v, w (that is, declare them in your Matlab script) so that a general solution is of the form v + aw, where a is any real number.