Exercise 4.5.11

Since $\delta$ is $n$-linear, we must have $\delta (A)=0$ if $A$ contains one zero row. Now if $M$ has rank less than $n$, we know that the $n$ row vectors of $M$ are dependent, say $u_1,u_2,\dots ,u_n$. So we can find some vector $u_i$ who is a linear combination of other vectors. We write

By Corollary 1 after Theorem 4.10 we can add $-a_j$ times the $j$-th row to the $i$-th row without changing the value of $\delta$. Let $M^{\prime }$ be the matrix obtained from $M$ by doing this processes. We know $M^{\prime }$ has one zero row, the $i$-th row, and hence $\delta (M_{\delta }(M^{\prime })=0$.

We can obtain $E_1$ from $I$ by interchanging two rows. By Theorem 4.10(a) we know that $\delta (E_1)=-\delta (I)$. Similarly we can obtain $E_2$ from $I$ by multiplying one row by a scalar $k$. Since $\delta$ is $n$-linear we know that $\delta (E_2)=\mathit{k\delta }(I)$. Finally, we can obtain $E_3$ by adding $k$ times the $i$-th row to the $j$-th row. By Corollary 1 after Theorem 4.10 we know that $\delta (E_3)=\delta (I)$.