Question

Prove det(AB)=(det A)(det B) for any nxn matrices A and B.

Answer 1

This is kind of tricky... I don't know if this answer is satisfactory, but this is basically true by definition.

If you have to be more detailed, it's going to be inconvenient... Since determinant is defined recursively, you could take an example, say a 2x2 matrix, and do this with letters:

Left side:
\[\det(\left[\begin{matrix}A & B \\ C & D\end{matrix}\right]) = AD - CB\]
\[\det(\left[\begin{matrix}E & F \\ G & H\end{matrix}\right]) = EH - FG\]
\[(AD - CB)(EH - FG) =\]

Right side:
\[\left[\begin{matrix}A & B \\ C & D\end{matrix}\right]\left[\begin{matrix}E & F \\ G & H\end{matrix}\right] = \left[\begin{matrix}AE + BG & AF + BH \\ CE + DG & CF + DH\end{matrix}\right]\]
\[\det(\left[\begin{matrix}AE + BG & AF + BH \\ CE + DG & CF + DH\end{matrix}\right]) = (AE + BG)(CF + DH) - (CE+DG)(AF + BH)\]

Then if you work everything else out to completion, they should give you the same thing.

Answer 2

This is very standard result and any linear algebra text worth its salt will have it. I'm not going to write it out as it's a pain and involves lots of alternating sums.

Answer 3

So, I figure I just may as well finish the problem. Also, my recommendation to you is to keep all the letters in alphabetical order so its easy to see what's going to cancel out.

\[(AD - CB)(EH - FG) = ADEH - CBEH - ADFG + CBFG\]

\[(AE + BG)(CF + DH) - (CE + DG)(AF + BH) = AECF + CBFG + ADEH\]
\[+ BGDH - [ AECF + ADGF + CBEH + BGDH]\] (continued from previous line)

So, distribute the negative sign through, and cancel stuff out, and you get:
\[ADEH - CBEH - ADFG + CBFG = ADEH - CBEH - ADFG + CBFG\]