${E}_{ij}(\lambda )$ is invertible with inverse ${E}_{ij}(\lambda )$ and ${E}_{i}(\lambda )$ is invertible with inverse ${E}_{i}(1/\lambda )$ (recall that $\lambda \ne 0$ for type II elementary matrices).
23. Elementary matrices, 2
23. Elementary matrices, 2
We'll prove it for 2by2 matrices to save ourselves from all the dots in the last few proofs. For the first: $${E}_{12}(\lambda ){E}_{12}(\lambda )=\left(\begin{array}{cc}\hfill 1\hfill & \hfill \lambda \hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 1\hfill & \hfill \lambda \hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill \lambda \lambda \hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)=I.$$ For the second: $${E}_{1}(\lambda ){E}_{1}(1/\lambda )=\left(\begin{array}{cc}\hfill \lambda \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 1/\lambda \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill \lambda /\lambda \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)=I.$$
We now proceed to the proof of the theorem:
If $A$ is a matrix and you use row operations on $(AI)$ until you get $(BC)$ with $B$ in reduced echelon form then:

if $B$ is $I$ then $A$ is invertible with inverse $C$ ;

if $B\ne I$ then $A$ is not invertible.
Put $A$ into reduced echelon form by performing row operations. Each row operation corresponds to multiplying $A$ on the left by an elementary matrix. Therefore $B={M}_{k}\mathrm{\cdots}{M}_{1}A$ for some sequence of elementary matrices ${M}_{i}$ .
We're simultaneously performing row operations on both sides of the bar in the augmented matrix, so $C={M}_{k}\mathrm{\cdots}{M}_{1}I={M}_{k}\mathrm{\cdots}{M}_{1}$ , so $C$ is a product of elementary matrices. Moreover, $B={M}_{k}\mathrm{\cdots}{M}_{1}A=CA$ .
If $B=I$ then $CA=I$ and $A$ is invertible with ${A}^{1}=C$ .
If $B\ne I$ , it needs to have a row of zeros. Indeed, suppose $B$ is an $n$ by$n$ matrix in echelon form. If all rows are nonzero then there are $n$ leading entries, which have to move to the right as you go down, so they have to live on the diagonal, which forces your matrix to be the identity.
Therefore there is a free variable in the solution to $Bv=0$ (equivalently $Av=0$ ). Therefore there is a whole line's worth of solutions, so there is at least one $v\ne 0$ such that $Av=0$ . If $A$ were invertible then $v={A}^{1}0=0$ , which is a contradiction. Therefore $A$ is not invertible.
A product of elementary matrices is invertible; conversely, any invertible matrix is a product of elementary matrices.
Recall that elementary matrices are invertible and their inverses are again elementary matrices.
If ${M}_{k}\mathrm{\cdots}{M}_{1}$ is a product of elementary matrices then ${({M}_{k}\mathrm{\cdots}{M}_{1})}^{1}={M}_{1}^{1}\mathrm{\cdots}{M}_{k}^{1}$ , so it is invertible. Conversely, if $A$ is invertible then its inverse has the form ${M}_{k}\mathrm{\cdots}{M}_{1}$ for some sequence of elementary matrices (as we just saw in the previous proof). Therefore $A={M}_{1}^{1}\mathrm{\cdots}{M}_{k}^{1}$ is a product of elementary matrices.