# Determinant of an elementary matrix

I read a very slick proof of determinant properties, in this case of the fact $\det A = \det A^T$, which says in one place

It suffices to notice that for any elementary matrix $M$ we have $\det M = \det M^T$

But it's not obvious to me. How can we show it?

Note: we're using the Laplace expansion for rows as the definition. It was noted that for elementary matrix $M$ we have $\det M = 1$ if $M$ adds a multiplied row, $\det M = -1$ if $M$ swaps two rows and $\det M = c$ if $M$ multiplies one row by $c$.

• Try writing each elementary operation as a matrix, then take its determinant and transpose. – vadim123 Feb 9 '16 at 21:52

1) $M$ multiplies a row by $c$
Then $M$ looks like $$\left[ \begin{matrix} 1 & & \\ & 1 \\ & & \ddots \\ & & & 1 \\ & & & & c \\ & & & & & 1 \\ & & & &&&\ddots \\ & & & &&&& 1 \end{matrix} \right]$$ So trivally $M^T = M$, hence the equality of deteminants
2) $M$ adds another row. Then the $M$ looks like $$\left[ \begin{matrix} 1 & & \\ & 1 & \ldots & a \\ & & \ddots & \vdots \\ & & & 1 \\ & & & &\ddots \\ & & & && 1 \end{matrix} \right]$$ By doing the Laplace expansion at the column containing $a$, we get $$detM = (-1)^{i+j} a \det I + (-1)^{j+j} \det I$$. This is symmetrical wrt to $i,j$, so in this case $\det M = \det M^T$ too
3) $M$ swaps two rows. Then $M$ looks like $$\left[ \begin{matrix} 1 & & \\ & \ddots \\ & & 1 \\ & & & 0 & & & & 1 \\ & & & & 1 \\ & & && & \ddots \\ & & & & & & 1 \\ & & & 1 & & & & 0 \\ & & & & & & & & 1 \\ & & & & & & & & & \ddots \\ & & & & & &&& &&1 \end{matrix} \right]$$ This matrix is symmetrical, so $M = M^T$, hence the equality of determinants.
• Your second case is wrong, In (2) it's not necessarily the case that $\tilde M_{ij}=I$. – Postal Model May 8 '18 at 8:12
• Say $M=I_n$, then $\tilde M_{ij}=I_{n-1}$ iff $i=j$. And albeit you're wrong at this part your result is correct because the $\det(\tilde{I_{ij}})=0$ when $i\not=j$ so $(-1)^{i+j} a \det(\tilde{I_{ij}})$ is zero. – Postal Model May 8 '18 at 8:26