as and so on.

- the row of as the row of
- the other rows of are the same as that of

Now,

Thus, Since, Therefore, has a right inverse. Hence, by Theorem 2.5.9 has an inverse and

height6pt width 6pt depth 0pt

This means, is invertible. Therefore, either is an elementary matrix or is a product of elementary matrices (see Theorem 2.5.8). So, let be elementary matrices such that Then, by using Parts 1, 2 and 4 of Lemma 2.6.6 repeatedly, we get

Thus, we get the required result in case is non-singular.

**Step 2.** Suppose

Then
is not invertible.
Hence, there exists an invertible matrix
such that
where
So,
and therefore

Thus, the proof of the theorem is complete. height6pt width 6pt depth 0pt

Suppose has an inverse. Then there exists a matrix such that Taking determinant of both sides, we get

This implies that Thus, is non-singular. height6pt width 6pt depth 0pt

If is singular, then Hence, by Corollary 2.6.16, doesn't have an inverse. Therefore, also doesn't have an inverse (for if has an inverse then Thus again by Corollary 2.6.16, Therefore, we again have

Hence, we have height6pt width 6pt depth 0pt

A K Lal 2007-09-12