3Elementary row operations and their …

1 3 Elementary row operations and their correspondingmatricesAs we ll see, any elementary row operation can be performed by multiplying the augmentedmatrix (A|y) on theleftby what we ll call anelementary matrix. Just so this doesn tcome as a total shock, let s look at some simple matrix operations : SupposeEAis defined, and suppose the first row ofEis (1,0,0, .. ,0). Then the firstrow ofEAisidenticalto the first row ofA. Similarly, if theithrow ofEis all zeros except for a 1 in theithslot, then theithrowof the productEAis identical to theithrow ofA. It follows that if we want tochange onlyrow i of the matrixA, we should multiplyAon the left by some matrixEwith the following property:Every rowexceptrow i should be theithrow of the corresponding identity procedure that we illustrate below is used to reduceanymatrix to echelon form (notjust augmented matrices ).

2 The way it works is simple: the elementary matricesE1, E2, ..are formed by (a) doing the necessary row operation on the identity matrix to getE, andthen (b) multiplyingAon the left : LetA=(34 52 1 0).1. To multiply the first row ofAby 1/3, we can multiplyAon the left by the elementarymatrixE1=(1300 1).(Since we don t want to change the second row ofA, the second row ofE1is the sameas the second row ofI2.) The first row is obtained by multiplying the first row ofIby1/3. The result isE1A=(143532 1 0).You should check this on your own. Same with the remaining To add -2(row1) to row 2 in the resulting matrix, multiply it byE2=(1 0 2 1).1 The general rule here is the following:To perform an elementary row operationon the matrixA, first perform the operation on the corresponding identitymatrix to obtain an elementary matrix; then multiplyAon the left by thiselementary with the problem, we obtainE2E1A=(143530 113 103).

3 Note the order of the factors:E2E1 Aand notE1E2A!3. Now multiply row 2 ofE2E1 Aby 3/11 using the matrixE3=(100 311),yielding the echelon formE3E2E1A=(143530 11011).4. Last, we clean out the second column by adding (-4/3)(row 2) to row 1. The corre-sponding elementary matrix isE4=(1 4301).Carrying out the multiplication, we obtain the Gauss-Jordan form of the augmentedmatrixE4E3E2E1A=(1 05110 11011).Naturally, we get the same result as before, so why bother? The answer is that we redeveloping an algorithm that will work in the general case. So it s about time to formallyidentify our goal in the general case. We begin with some definitions. Definition:Theleading entryof a matrix row is the first non-zero entry in the row,starting from the left.

4 A row without a leading entry is a row of zeros. Definition:The matrixRis said to be inechelon formprovided that1. The leading entry of every non-zero row is a If the leading entry of rowiis in positionk, and the next row is not a row of zeros,then the leading entry of rowi+ 1 is in positionk+j, wherej All zero rows are at the bottom of the following matrices are in echelon form:(1 0 1), 1 0 0 10 0 0 ,and 0 1 0 0 1 0 0 0 1 .Here the asterisks (*) stand for any number at all, including 0. Definition:The matrixRis said to be inreduced echelon formif (a)Ris in echelonform, and (b) each leading entry is theonlynon-zero entry in its column. The reducedechelon form of a matrix is also called following matrices are in reduced row echelon form:(1 00 1), 1 0 0 0 1 0 0 0 0 ,and 0 1 0 0 0 0 1 0 0 0 0 1.

5 Exercise: SupposeAis 3 5. What is the maximum number of leading 1 s that can appearwhen it s been reduced to echelon form? Same questions forA5 3. Can you generalize yourresults to a statement forAm n?. (State it as a theorem.)Once a matrix has been brought to echelon form, it can be put into reduced echelon formby cleaning out the non-zero entries in any column containing a leading 1. For example, ifR= 1 2 1 30 12 00 00 1 ,which is in echelon form, then it can be reduced to Gauss-Jordan form by adding (-2)(row2) to row 1, and then (-3)(row 3) to row 1. Thus 1 2 001 000 1 1 2 1 30 12 00 00 1 = 1 0 5 30 12 00 00 1 .and 1 0 30 100 01 1 0 5 30 12 00 00 1 = 1 0 5 00 12 00 00 1.

6 Note that column 3 cannot be cleaned out since there s no leading 1 is one more elementary row operation and corresponding elementary matrix we mayneed. Suppose we want to reduce the following matrix to Gauss-Jordan form3A= 22 10031 12 .Multiplying row 1 by 1/2, and then adding -row 1 to row 3 leads toE2E1A= 1 0 00 1 0 1 0 1 120 00 1 00 0 1 22 10031 12 = 11 120030 252 .Now we can clearly do 2 more operations to get a leading 1 in the (2,3) position, and anotherleading 1 in the (3,2) position. But this won t be in echelon form (why not?) We need tointerchange rows 2 and 3. This corresponds to changing the order of the equations, andevidently doesn t change the solutions.

7 We can accomplish this by multiplying on the leftwith a matrix obtained fromIby interchanging rows 2 and 3:E3E2E1A= 1 0 00 0 10 1 0 11 120030 252 = 11 120 252003 .Exercise: Without doing any written computation, write down the Gauss-Jordan form forthis : Use elementary matrices to reduceA=(2 1 1 3)to Gauss-Jordan form. You should wind up with an expression of the formEk E2E1A= is another name for the matrixB=Ek E2E1?4