Math 1b Practical — Basic matrices and solutions; pivot operationsJanuary 3, 2011We use matrices to model systems of linear equations. For example, the system2x1− 3x2− 5x3− 7x4+2x5=7x1+ x2− 5x3+ x4− 4x5=3x1− 4x3+3x4+ x5= −4(1)has corresponding matrix2 −3 −5 −7211−51−410−43 173−4. (2)Two matrices (of the same dimensions) are row-equivalent when one can be obtainedfrom the other by a sequence of elementary row operations as described in LADW, Sec-tion 2.1 on page 40. It is important to understand that the systems of linear equationscorresponding to two row equivalent matrices will have exactly the same set of solutions.An r by n matrix M with all nonzero rows is basic when its columns include the unitvectors⎛⎜⎜⎜⎜⎝100...0⎞⎟⎟⎟⎟⎠,⎛⎜⎜⎜⎜⎝010...0⎞⎟⎟⎟⎟⎠,⎛⎜⎜⎜⎜⎝001...0⎞⎟⎟⎟⎟⎠, ...,⎛⎜⎜⎜⎜⎝000...1⎞⎟⎟⎟⎟⎠(of height r)(3)in some order. These may be called the special columns of M. A matrix with some rowsof all zeros is called basic when the rows of all zeros are at the bottom and the submatrixconsisting of the nonzero rows is basic.All matrices in reduced echelon form are basic matrices. Other examples of e.g. 3 × 7basic matrices are⎛⎝230410567180095505015⎞⎠,⎛⎝217410560884190000000⎞⎠,⎛⎝210410060180015007010⎞⎠.In the third matrix, there are several choices for the columns that contain the vectors in(3). In such a case, one should have a particular choice of special columns in mind. Forexample, in the right-most matrix above, we may choose to regard columns 3, 5, and 6 asthe special columns, though there is nothing wrong with another choice like 2, 6, 7.To solve a system of linear equations, we start with the matrix M =A bofthe system (here A is the matrix of coefficients of the variables and b is the column ofscalars on the right hand side of the equations) and use row operations until we find arow-equivalent matrix M=Abin which Ais basic. If the indices of the columnsof Acontaining the vectors in (3) are j1,j2,...,jr, then the variables xj1,,xj2,,...,xjrin the new, equivalent system have been isolated in the sense that each appears withnonzero coefficient in only one of the equations. The variables xj1,,xj2,,...,xjrmay becalled “dependent” variables, since their values are determined uniquely by the values ofthe others, which may be called “free” variables.[Caution: Which variables are free and which are dependent depends on the basicform. This is not determined by the original equations.]For example (see the last page), the matrix in (2) is row-equivalent to0 001 15/19−1/4100−117/38−1/4010 13/38−26/1933/19−10/19.The corresponding system of linear equations isx4+(15/19)x5= −26/19(−1/4)x1+ x2− (117/38)x5=33/19(−1/4)x1+ x3+(13/38)x5= −10/19 ,or, what is the same,x4= −(26/19) − (15/19)x5x2=33/19 + (1/4)x1+ (117/38)x5x3= −(10/19) + (1/4)x1− (13/38)x5.(4)Here x2,x3,x4are the dependent variables and x1,x5the free variables. The originalsystem of equations may now be considered as “solved” because we have described allsolutions in a simple way, in terms of two “parameters”.If we set the free variables to 0, we obtain a particular solution of the system:x1=0,x2=33/19,x3= −10/19,x4= −26/19,x5=0.This is an example of a basic solution.In general, if the coefficient matrix (not including the last column of a nonhomogeneoussystem) of a system of linear equations is basic, and the free variables are assigned thevalue 0, then the resulting solution of the system is called a basic solution.Thevaluesofthe dependent variables will be, in some order, the numbers in the last column. There willin general be many basic solutions to a system, corresponding to different basic matricesrow-equivalent to the original matrix. Exercise: What are the basic solutions of the systemabove corresponding to the basic matrices on the last page of these notes?[Caution: The usage of the terms basic matrix and basic solution is not completelystandard.]Notice that if a system of linear equations has a solution (at least one), then it has abasic solution. One explanation is simply that we can see whether a system has a solutionby looking at the reduced echelon (or any basic) form, and if the system does have asolution, a basic solution is evident. We will clarify this in class along with the assertionthat a system of equations has only finitely many basic solutions.***It is convenient to combine some elementary row operations into a single operationwe call a pivot operation.IfM =⎛⎜⎜⎝a11a12... a1na21a22... a2n...... ......am1am2... amn⎞⎟⎟⎠and aij= 0, we may pivot on position (i, j) by, first, multiplying row i by 1/aijso thatthe (i, j) position becomes 1, and then adding (or subtracting) multiples of the new rowi from the other rows so that all entries of column j, except the entry in position (i, j)become 0.For example, if we pivot on position (1, 1) of a 3 by 2 matrix with a11=0,weget⎛⎝1 a21/a110 a22− a21a12/a110 a32− a31a12/a11⎞⎠The next page shows Mathematica code for a function pivot which, when given M,i, j, returns the matrix obtained from M by pivoting on position (i, j). Warning: Do nottry this when the (i, j)-entry is 0. Several examples are given.pivot@M_,i_,j_D :=Block@8m<,m= M; m@@iDD = m@@iDD ê m@@i, jDD;Do@If@k  i, , m@@kDD = m@@kDD − m@@k, jDD m@@iDDD, 8k, Length@mD<D;mDMatrixForm@A = 882, −3, −5, −7, 2, 7<, 81, 1, −5, 1, −4, 3<, 81, 0, −4, 3, 1, −2<<D2 −3 −5 −72 711 −51 −4310 −43 1 −2MatrixForm@A = RowReduce@ADD10−40−2619401901−10−65194319000 11519−2619MatrixForm@A = pivot@A, 1, 5DD−19260381301−2013−5219 00−315260 −301310−213MatrixForm@A = pivot@A, 3, 3DD00019151 −2615−141039100 −185−1401−13300115MatrixForm@A = pivot@A, 1,