Answer
We prove:
$\det \left( {AB} \right) = \det \left( A \right)\det \left( B \right)$
Work Step by Step
Let the matrices $A$ and $B$ be given by:
$A = \left( {\begin{array}{*{20}{c}}
a&b\\
c&d
\end{array}} \right)$, ${\ \ \ \ \ }$ $B = \left( {\begin{array}{*{20}{c}}
{a'}&{b'}\\
{c'}&{d'}
\end{array}} \right)$
So,
$AB = \left( {\begin{array}{*{20}{c}}
a&b\\
c&d
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{a'}&{b'}\\
{c'}&{d'}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{aa' + bc'}&{ab' + bd'}\\
{ca' + dc'}&{cb' + dd'}
\end{array}} \right)$
By Eq. (1) in Section 13.4, the determinant of $AB$ is
$\det \left( {AB} \right) = \left( {aa' + bc'} \right)\left( {cb' + dd'} \right) - \left( {ca' + dc'} \right)\left( {ab' + bd'} \right)$
$ = aca'b' + bcb'c' + ada'd' + bdc'd'$
${\ \ \ }$ $ - aca'b' - adb'c' - bca'd' - bdc'd'$
After canceling terms with opposite signs, we get
$\det \left( {AB} \right) = bcb'c' + ada'd' - adb'c' - bca'd'$
$ = bc\left( {b'c' - a'd'} \right) + ad\left( {a'd' - b'c'} \right)$
$ = ad\left( {a'd' - b'c'} \right) - bc\left( {a'd' - b'c'} \right)$
$ = \left( {ad - bc} \right)\left( {a'd' - b'c'} \right)$
Since $\det A = ad - bc$ and $\det B = a'd' - b'c'$, hence
$\det \left( {AB} \right) = \det \left( A \right)\det \left( B \right)$
