Math 1b Practical — Problem Set 1Due 3:00pm, Monday, January 10, 2011See the handout “Basic solutions; pivot operations” for the definition of pivot operationand basic solution in case you start this problem set before the terms are defined in class.1. (40 points) Let the matrixA =⎛⎝2 −13 5−320−1−4333⎞⎠be the coefficient matrix of a homogeneous system of three linear equations in four variablesw, x, y, z.(i) Use pivot operations (and interchange of rows, if necessary) to find the reducedechelon form Aof A. (Show the results after each pivot.) Describe the solutions of thesystem by writing some variables in terms of the others (in terms of the “free” variableswith respect to A).(ii) Pivot on position (1, 4) in the fourth column of Ato find another row-equivalentbasic matrix A. Then describe the solutions of the original system by writing some of thevariables in terms of the “free” variables with respect to A.(iii) Consider the three (nonhomogeneous) systems of linear equations Ax = b for thefollowing values of b:⎛⎝123⎞⎠,⎛⎝001⎞⎠,⎛⎝125⎞⎠.In each case, start with [A, b] and perform pivot operations in the first four columns sothat the initial 3 × 4 submatrix of the result is in reduced echelon form. (So you will haveto repeat the pivots you did in part (i), but see the suggestion below, first).Describe the solutions of each system. In particular, decide which are consistent. Forthose systems that are consistent, give the basic solution of the system corresponding toA.Suggestion: For efficiency, I suggest doing all three problems at the same time. Startwith the 4 × 7matrixM =A B=⎛⎝2 −135101−320−1202−4333315⎞⎠,where A is as above and B is the 3 × 3 matrix whose columns are the three choices of b.Do pivot operations in the first four columns to obtain a new matrixM=ABwhere Ais in reduced echelon form, as in part (ii). Show M, but don’t show any inter-mediate steps.(iv) Do one more pivot on M, in position (1, 4), to getM=ABand show M. For those systems that are consistent, give the new basic solutions of thesystem corresponding to A.[Check some of your work, i.e. make sure your basic solutions are solutions of theoriginal system. But there is no need to write this down when you turn in your work.]2. (25 points)(i) (23 points) Consider the system of equationsa11x1+ a12x2+ a13x3= b1,a21x1+ a22x2+ a23x3= b2,with matrixA =a11a12a13b1a21a22a23b2.Pivot on position (1, 1) of A to get a matrix A(show A) and and then on position (2, 2)of Ato get a matrix A(show A) corresponding to an equivalent system of equations ofthe formx1+ px3= s,x2+ qx3= t.I want to know what p, q, s, t are. [This is somewhat grungy.][This calculation works if the coefficients aijare symbols or variables, but won’t bevalid in some cases if they are numbers, because we aren’t allowed to pivot on a positioncontaining 0.](ii) (2 points). Now imagine a system of 20 linear equations in 21 variables, withgeneral coefficients aijand bj. [In practical applications, this is not a large system atall.] Imagine that you have pivoted on positions (1, 1), (2, 2), up to (20, 20) to solve forx1,x2,...,x20in terms of x21and the coefficients. Guess how many pages it would taketo write down the formulas for the answers, in this general form. [Don’t spend a lot oftime on this. You will get the 2 points unless your answer is very very very wrong, like 3pages or 101000pages. You may give your answer in terms of gigabytes if you prefer.]3. (15 points) Explain why (i.e. prove that) if we perform a pivot operation on the positionof any nonzero entry of a basic matrix, the resulting matrix is again basic.4. (20 points) LetA =⎛⎜⎝a11a12... aana21a22... a2na31a32... a3na41a42... a4n⎞⎟⎠be a matrix with four rows. LetA=⎛⎜⎝• 0 • ... •• 0 • ... •• 1 • ... •• 0 • ... •⎞⎟⎠be the matrix obtained from A by a pivot operation in position (3, 2), assuming a32=0.(i) Explicity give (in terms of the entries aij)the4by4matrixU so that A= UA.(ii) What is