2.1. MATRICES 49

and if A−1 existed, this could not happen because you could multiply on the left by theinverse A and conclude the vector (−1, 1)

T= (0, 0)

T. Thus the answer is that A does not

have an inverse.Suppose you want to find B such that AB = I. Let

B =(

b1 · · · bn

)Also the ith column of I is

ei =(

0 · · · 0 1 0 · · · 0)T

Thus, if AB = I, bi, the ith column of B must satisfy the equation Abi = ei. The augmented

matrix for finding bi is (A|ei) . Thus, by doing row operations till A becomes I, you end upwith (I|bi) where bi is the solution to Abi = ei. Now the same sequence of row operationsworks regardless of the right side of the agumented matrix (A|ei) and so you can save troubleby simply doing the following.

(A|I) row operations→ (I|B)

and the ith column of B is bi, the solution to Abi = ei. Thus AB = I.This is the reason for the following simple procedure for finding the inverse of a matrix.

This procedure is called the Gauss Jordan procedure. It produces the inverse if the matrixhas one. Actually, it produces the right inverse.

Procedure 2.1.25 Suppose A is an n × n matrix. To find A−1 if it exists, form theaugmented n× 2n matrix,

(A|I)

and then do row operations until you obtain an n× 2n matrix of the form

(I|B) (2.18)

if possible. When this has been done, B = A−1. The matrix A has an inverse exactly whenit is possible to do row operations and end up with one like 2.18.

As described above, the following is a description of what you have just done.

ARqRq−1···R1→ I

IRqRq−1···R1→ B

where those Ri sympolize row operations. It follows that you could undo what you did bydoing the inverse of these row operations in the opposite order. Thus

IR−1

1 ···R−1q−1R

−1q→ A

BR−1

1 ···R−1q−1R

−1q→ I

Here R−1 is the row operation which undoes the row operation R. Therefore, if you form(B|I) and do the inverse of the row operations which produced I from A in the reverseorder, you would obtain (I|A) . By the same reasoning above, it follows that A is a rightinverse of B and so BA = I also. It follows from Proposition 2.1.23 that B = A−1. Thusthe procedure produces the inverse whenever it works.