10.5. JUSTIFICATION FOR THE MULTIPLIER METHOD 209

Lemma 10.5.1 Let L be a lower (upper) triangular matrix m×m which has ones down themain diagonal. Then L−1 also is a lower (upper) triangular matrix which has ones downthe main diagonal. In the case that L is of the form

L =

1a1 1...

. . .

an 1

 (10.1)

where all entries are zero except for the left column and main diagonal, it is also the casethat L−1 is obtained from L by simply multiplying each entry below the main diagonal in Lwith −1. The same is true if the single nonzero column is in another position.

Proof: Consider the usual setup for finding the inverse(

L I). Then each row oper-

ation done to L to reduce to row reduced echelon form results in changing only the entries inI below the main diagonal. In the special case of L given in 10.1 or the single nonzero col-umn is in another position, multiplication by −1 as described in the lemma clearly resultsin L−1 . ■

For a simple illustration of the last claim, 1 0 0 1 0 00 1 0 0 1 00 a 1 0 0 1

→ 1 0 0 1 0 0

0 1 0 0 1 00 0 1 0 −a 1

Now let A be an m×n matrix, say

A =

a11 a12 · · · a1n

a21 a22 · · · a2n...

......

am1 am2 · · · amn

and assume A can be row reduced to an upper triangular form using only row operation 3.Thus, in particular, a11 ̸= 0. Multiply on the left by E1 =

1 0 · · · 0− a21

a111 · · · 0

......

. . ....

− am1a11

0 · · · 1

This is the product of elementary matrices which make modifications in the first columnonly. It is equivalent to taking −a21/a11 times the first row and adding to the second. Thentaking −a31/a11 times the first row and adding to the third and so forth. The quotients inthe first column of the above matrix are the multipliers. Thus the result is of the form

E1A =

a11 a12 · · · a′1n

0 a′22 · · · a′2n...

......

0 a′m2 · · · a′mn

