Exercise 4.5.12

If $A$ is not full-rank, we have that $\mathit{AB}$ will be not full-rank. By Corollary 3 after Theorem 4.10 we have

If $A$ is full-rank and so invertible, we can write $A=E_s\cdots E_2{E}_1$ as product of elementary matrices. Assuming the fact, which we will prove later, that

for all elementary matrix $E$ and all matrix $M$ holds, we would have done since

So now we prove the fact. First, if $E$ is the elementary matrix of type 1 meaning interchanging the $i$-th and the $j$-th rows, we have $\mathit{EM}$ is the matrix obtained from $M$ by interchanging the $i$-th and the $j$-th rows. By Theorem 4.10(a) we know that

Second, if $E$ is the elementary matrix of type 2 meaning multiplying the $i$-th row by a scalar $k$, we have $\mathit{EM}$ is the matrix obtained from $M$ by multiplying the $i$-th row by scalar $k$. Since the function $\delta$ is $n$-linear, we have

Finally, if $E$ is the elementary matrix of type 3 meaning adding $k$ times the $i$-th row to the $j$-th row, we have $\mathit{EM}$ is the matrix obtained from $M$ by adding $k$ times the $i$-th row to the $j$-th row. By Corollary 1 after Theorem 4.10, we have

This completes the proof.