There are three types of elementary matrices:

  • Type 1: matrices obtained by interchanging the ith row of $I$ and jth row of $I$;

  • Type 2: matrices obtained by multiplying the ith row of $I$ by $\lambda\neq 0$;

  • Type 3: matrices obtained by adding $\lambda$ times the jth row of $I$ to the ith row of $I$.

I know that every invertible matrix $A$ can be written as a product of elementary matrices (of types 1, 2 and 3). According to Example 12.2.5 here, it is clear that

(a) if $\det A=1$ then $A$ can be written as a product of elementary matrices of type 2 and type 3.

(b) if $\det A=1$ and the factorization of $A$ (in terms of elementary matrices) has a matrix of type 2 with some $\lambda$, than this factorization also has a matrix of type 2 with $\lambda^{-1}$.

The item (b) is true because otherwise we would have $\det A\neq 1$. However, I can't understand why (a) is true. So, I'd like help to prove it.




the determinat of a product of matrices is the product of the determinants, and an elementary matrix of type 1) has negative determinat (it is an alternating multilinear form).

  • $\begingroup$ For me, the first implication of your hint is: the number $k$ of matrices of type 1 in the factorization of $A$ is even. Why $k>0$ is not possible? $\endgroup$ – Pedro May 17 '15 at 23:58

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