17
Feb 19

## Determinant of a product

The derivation is very similar to the proof of the Leibniz rule.

Consider matrices $A,B$ and let $B_j$ denote the $j$th row of $B$. Row $C_i$ of the product $C=AB,$ obviously, can be written as $C_i=\sum_{j=1}^na_{ij}B_j.$ Using Property V $n$ times, we have

(1) $\det C=\det \left(\begin{array}{c}\sum_{j_1=1}^na_{1j_1}B_{j_1} \\ C_2 \\... \\ C_n\end{array}\right) =\sum_{j_1=1}^na_{1j_1}\det \left(\begin{array}{c}B_{j_1} \\ C_2 \\ ... \\ C_n\end{array}\right)$ $=\sum_{j_1=1}^na_{1j_1}\det \left(\begin{array}{c}B_{j_1} \\ \sum_{j_2=1}^na_{2j_2}B_{j_2} \\ ... \\ C_n\end{array}\right)=\sum_{j_1,j_2=1}^na_{1j_1}a_{2j_2}\det\left(\begin{array}{c}B_{j_1} \\ B_{j_2} \\ ... \\ C_{n}\end{array}\right)$ $=...=\sum_{j_1,...,j_n=1}^na_{1j_1}...a_{nj_n}\det\left(\begin{array}{c} B_{j_1} \\ B_{j_2} \\ ... \\ B_{j_n}\end{array}\right).$

By analogy with permutation matrices denote $B_{j_1,...,j_n}=\left(\begin{array}{c}B_{j_1} \\B_{j_2} \\... \\B_{j_n}\end{array}\right).$

Recall the link between $B_{j_1,...,j_n}$ and $B:$

(2) $B_{j_1,...,j_n}=P_{j_1,...,j_n}B.$

If we had proved the multiplication rule for (2), we would have

(3) $\det B_{j_1,...,j_n}=(\det P_{j_1,...,j_n})(\det B).$

In the theory of permutations (3) is proved without relying on the multiplication rule. I am going to use (3) as a shortcut that explains the idea. Combining (1) and (3) we obtain by the Leibniz formula $\det C=\sum_{j_1,...,j_n:\det P_{j_1,...,j_n}\neq 0}a_{1j_1}...a_{nj_n}(\det P_{j_1,...,j_n})(\det B)=(\det A)(\det B).$